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College Algebra and Trigonometry Alexander Rozenblyum

New York City College of Technology, CUNY

Copyright © 2018 Alexander Rozenblyum

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Preface

This textbook is based on the course “College Algebra and Trigonometry” taught at New

York City College of Technology, CUNY. It is designed to prepare students for the

precalculus class at the next level.

The book is meant to be concise, while at the same time including all the material taught

at the present time (as of Fall 2020) by mathematics department of NYCCT. Special

attention is paid to present the material in a motivated and intuitively clear manner.

The book is divided into three parts and 25 sessions based on the total of 30 sessions in

one semester, leaving remaining 5 sessions for 3 in-class exams, one review for final

exam, and one final exam.

Each session ends with exercises. Most exercises are designed in pairs with consecutive

odd and even numbers in such a way that exercises are similar in each pair. A

recommended approach is for instructors to work out the examples in the lecture part of

the session during class, then allow students to work on even exercises during class and

assign the odd exercises for homework (or vice versa). The textbook also contains

answers to almost all exercises.

Some sessions contain challenge problems. In sessions 4, 5, 8, 13 and 23, challenge

problems are given in parametric form, so they also can be used by instructors to generate

additional specific exercises and problems for quizzes and exams.

I would like to thank my colleagues from the Mathematics Department. My special

thanks to Thomas Tradler and Holly Carley for a careful reading of the first part of the

book and many valuable comments and suggestion. I am grateful to Roy Berglund for a

detailed review of the text and many useful tips. I thank Satyanand Singh and Joel

Greenstein for suggesting for improving the text.

Also, I would like to thank my brother, Dr. Leonid Rozenblum, for numerous corrections

and recommendations.

“Impossible triangle” on the title page is called the Penrose Triangle.

Alexander Rozenblyum

January, 2021

Contents

Part 1: Rational and Irrational Expressions and Equations …………………… 1

Session1: Systems of Three Linear Equations in Three Variables …. 2

Session 1A: Determinants and Cramer’s Rule (Optional) ……………. 8

Session2: Quadratic Equations: Factored Form ……………………. 16

Session 3: Integer Exponents ………………………………………... 22

Session 4: Rational Expressions and Complex Fractions …………… 30

Session 5: Rational Equations ………………………………………. 42

Session 6: Radicals and Fractional Exponents ……………………… 49

Session 7: Multiplication, Addition and Subtraction of Radical

Expressions ………………..……………………………... 57

Session 8: Rationalizing the Denominators and

Solving Radical Equations ……………………………….. 63

Session 9: Complex Numbers and Squared Form

of Quadratic Equations …………………………………... 70

Session 10: Completing the Square and the Quadratic Formula …..…. 78

Session 11: Parabolas …………………………………………………. 85

Session 12: Distance Formula, Midpoint Formula, and Circles ………. 97

Session 13: Nonlinear Systems of Equations in Two Variables ………. 104

Part 2: Trigonometric Functions ………………….111

Session 14: Geometric and Trigonometric Angles …………………..... 112

Session 15: Trigonometric Functions for Acute Angles …………….... 108

Session 16: Trigonometric Functions for Arbitrary Angles ………….. 127

Session 17: Solving Oblique Triangles – Law of Sines ………………. 137

Session 18: Solving Oblique Triangles – Law of Cosines ……………. 147

Session 19: Radian Measure of Angles ………………………………... 154

Session 20: Graph and Basic Function for Sine………………………… 161

Session 21: Graphs and Basic Functions for Cosine and Tangent……… 168

Session 22: Trigonometric Identities and None-Basic Equations ……... 175

Part 3: Exponential and Logarithmic Functions …………………. 181

Session 23: Logarithms ………………………………………………… 182

Session 24: Exponential and Logarithmic Functions ………………….. 190

Session 25: Compound Interest and Number e ………………………… 198

Answers to exercises …………………………………………………….. 205

Index …………………………………………………………………….. xxx

Part I

Rational and Irrational

Expressions and Equations

1

Session 1: Systems of Three Linear Equations in Three Variables

Session 1

Systems of Three Linear Equations in Three Variables

We assume that readers are already familiar with linear equations in one variable and

with systems of two linear equations in two variables. Linear equations in one variable x

can be written as ax + b = c, and systems of two linear equations in two variables x and y

can be written as

1 1 1

2 2 2

a x b y c

a x b y c

+ =

+ =.

In this session, we consider systems of three linear equations in three variables.

The general form of such systems is this

=++

=++

=++

3333

2222

1111

dzcybxa

dzcybxa

dzcybxa

.

Here x, y, and z are variables (unknown values). All other letters are given numbers.

Numbers that are written next to variables (labeled with letters a, b and c) are called the

coefficients of the system. The above system has 9 coefficients. A solution of the system

is a triple (X, Y, Z) that satisfies the system (makes each equation a true statement after

substituting these numerical values for variables (x, y, z)).

Note. Keep in mind that a triple (X, Y, Z) represents one solution, not three.

There are different methods of solving systems of linear equations with any number of

equations and any number of variables. Here we consider the elimination method. This

method suggests that we eliminate one of the variables from two equations of the syst em

using the third equation. After eliminating this variable, we get two equations with the

two other variables. We can solve this system using the elimination method again. As a

result, we will find the values of two unknowns. Finally, we substitute these values into

one of the equations of the original system and solve it for the third unknown. To

eliminate a variable, we multiply equations by appropriate numbers and then add them

up. For this reason, this method is also called the addition-elimination method.

Theoretically, there are three possibilities regarding the number of solutions of the linear

system: it may have

1) One solution (so, one triple). We also say that the system has unique solution.

2) No solutions at all. Such system is called inconsistent.

3) Infinite many solutions. Such system is called dependent.

In the examples below, we solve several systems of equations. In doing this, we often

deal with the movement of terms from one side of the equation to another. Technically,

this action can be performed directly by moving the terms and changing their signs (see

Example 1.0 below) or indirectly by adding / subtracting the same terms on both sides of

2


the equation. We call the direct method the Moving method, and the indirect method the

Adding method. As an example, let’s compare these methods when solving the equation

2x – 1 = x + 5.

Example 1.0. Solve the equation 2x – 1 = x + 5 with both methods: adding and moving.

To solve the equation, we collect the terms with unknown x on the left side of the

equation, and the numerical terms on the right side.

Adding Method Moving Method

Write +1 underneath both sides of the equation:

2x – 1 = x + 5

+ 1 + 1

Cancel –1 and 1 on the left side and add 5 and 1

on the right side:

2x = x +6

Write –x underneath both sides:

2x = x +6

–x –x

Subtract terms on the left side and cancel x on

the right side:

x = 6.

Move –1 to the right side and move x to the

left side, change signs of both to the

opposite:

2x – x = 5 + 1

Subtract terms on the left side and add

terms on the right side:

x = 6.

As you can see, the Moving method is simpler. In this textbook, we will most often use it.

Example 1.1. Solve the system

3 2 3

2 4 5 1

8 3 3 17

x y z

x y z

x y z

− + = −

+ − =− + + =

.

Solution. We have many options to eliminate variables. Actually, we can eliminate

anyone. Let’s eliminate y from the second and third equations using the first equation in

which the coefficient for y is –1. It reduces our work if we use a variable with a

coefficient of 1 or –1.

1) Consider the first and second equations:

=−+

−=+−

1542

323

zyx

zyx

To eliminate y, we want the coefficients for y in both equations to be equal in absolute

values but have the opposite signs. In this case, if we add the equations, variable y will be

cancelled (eliminated). To get this case, it’s enough to multiply the first equation by 4.

Note. Multiplication an equation by a number means multiplying all the terms of the

equation by that number. We get

3


=−+

−=+−

1542

128412

zyx

zyx

Now we add these equations and y is eliminated:

12x + 2x + 8z – 5z = – 12 + 1, or 14x + 3z = –11.

2) To eliminate y from the third equation, consider the first and third equations together:

=++−

−=+−

17338

323

zyx

zyx

Number 3 outside the braces means that we intend to multiply the first equation by 3:

=++−

−=+−

17338

9639

zyx

zyx

Add these equations to eliminate y: 9x –8x + 6z +3z = –9 + 17, or x + 9z = 8.

3) Combine the resulting equations from steps 1 and 2 into one system:

=+

−=+

89

11314

zx

zx.

4) Solve the above system (using the elimination method again):

=+

−=+

89

11314

zx

zx

−=−−

−=+

11212614

11314

zx

zx

3z – 126z = –11 – 112 –123z = – 123 z = 1.

5) At this point we found the variable z = 1. Now we move back in the above steps.

Substitute z = 1 into the second equation in step 3) and solve for x:

9 1 8 1.x x+ = = −

6) Substitute the values x = – 1 and z = 1 into the first equations of the original system,

and solve it for y:

312)1(3 −=+−− y , –3 –y + 2 = –3, –y = – 3 + 3 –2, –y = –2, y = 2.

Final answer: the system has one (unique) solution

x = –1, y = 2, z = 1, or as a triple (–1, 2, 1).


2 4 5

2 3 1

4 7 7

x y z

x y z

x y z

− + =

+ − = − + =

.

3

–14

4


Solution. Let’s eliminate z from the first and third equations using the second equation

(which has coefficient –1 for z). Of course, you may eliminate any other variable.

1) Consider the first and second equations:

2 4 5

2 3 1

x y z

x y z

− + =

+ − =

2 4 5

8 12 4 4

x y z

x y z

− + =

+ − =

Add the last equations to eliminate z: x + 8x – 2y + 12y = 5 + 4, 9x + 10y = 9.

2) Consider the second and third equations:

=+−

=−+

774

132

zyx

zyx

14 21 7 7

4 7 7

x y z

x y z

+ − =

− + =

Add the last equations: 14x + 4x +21y – y = 7 + 7, 18x + 20y = 14, 9x + 10y = 7.


9 10 9

9 10 7

x y

x y

+ =

+ =

4) Solve the above system. Notice that the left sides of both equations are the same but

the right sides are different. Therefore, this system does not have solutions, so the

system is inconsistent.

Final answer: the system does not have solutions. We can also say that the solution set

is an empty set (the symbol for empty set is ).


−=−+

=+−

=+−

423

7568

5342

zyx

zyx

zyx

.

Solution. Let’s eliminate x from the first and second equations using the third equation

(which has coefficient 1 for x).

1) Consider the first and third equations:

−=−+

=+−

423

5342

zyx

zyx

=+−−

=+−

8462

5342

zyx

zyx

Add the last equations to eliminate x: – 4y – 6y + 3z + 4z = 5 + 8, –10y + 7z = 13.

2) Consider the second and third equations:

−=−+

=+−

423

7568

zyx

zyx

=+−−

=+−

3216248

7568

zyx

zyx

4

7

–2

–8

5


Add the last equations: – 6y – 24y + 5z + 16z = 7 + 32, – 30y + 21z = 39,

–10y + 7z = 13.


=+−

=+−

13710

13710

zy

zy.

4) Solve the above system. Notice that both equations coincide. So, actually, we have

only one equation. In this case we cannot find the values of y and z uniquely. Indeed,

we can assign any numerical value to one of the variables y or z, say to z. Then we

can solve the above equation for y. Since there are infinite values of z to choose from,

we get an infinite number of pairs ),( zy which are solutions of the above equation. It

means that the system has infinitely many solutions. Substituting y and z into any of

the original equations, we can find x. Finally, we will get infinitely many triples

(X, Y, Z). So, the system is dependent.

We come up to an interesting question, how to describe an infinite set of all solutions of

the system. Of course, we cannot create an infinite list of them. Instead, we can use the

parametric form to describe the solution set. It means the following. Let’s solve the

above equation –10y + 7z = 13 for y in terms of z:

13 710 7 13 10 13 7 .

10 10y z y z y z− + = − = − = − +

Here the variable z may take any values, and we call it the free parameter. Let’s denote

this parameter by the letter t: z = t. Then, ty10

7

10

13+−= . Now, we can express the

variables x in terms of the parameter t by substituting expressions for y and z into any

equation of the original system. Let’s substitute expressions for y and z into the thirds

equation (in which the coefficient for x is 1) and solve for x:

13 7 39 213 2 4 2 4

10 10 10 10x t t x t t

+ − + − = − − + − = −

,

39 21 39 21 20 40 12 4

10 10 10 10

t t tx t t

− + − − −= − + − = = .

We can also write x as tx10

1

10

1−−= . Now we have described all unknowns in

parametric form:

1 1 13 7, , .

10 10 10 10x t y t z t= − − = − − =

Here t is a parameter that takes any numerical value.

Note. We can get the specific (particular) numerical solutions of the original system from

the above parametric representation by assigning any specific number to the parameter t.

For example, if we put t = 0, we get the particular solution 0,10

13,

10

1=−=−= zyx .

6


Exercises 1

In exercises 1.1 and 1.2, solve the system of equations.

If the system is inconsistent, state that.

If the system is dependent, state that.

1.1.

a)

3 6 4 1

5 2 8

6 6

x y z

y z

z

+ − =

+ = = −

b)

2 3 10

5 2 3 1

3 4 5 5

x y z

x y z

x y z

− + = −

+ − =− + + = −

c)

3 2 1

11 10 9 5

2 3 1

x y z

x y z

x y z

+ + =

+ + = + + =

d)

4 2 3 7

6 2 4

5 3 4 2

x y z

x y z

x y z

− + =

+ − = −− + + = −

e)

5 3 8 6

2 1

3 2 5 4

x y z

x y z

x y z

+ − =

+ − = + − =

1.2.

a)

4 3 2 4

2 5 19

3 9

x y z

y z

z

+ − =

− = − =

b)

4 2 3 10

2 5 21

8 7 5 6

x y z

x y z

x y z

+ − =− + + = − + − =

c)

3 4

3 2 3

6 4 2 1

x y z

x y z

x y z

− + =

+ + = + + = −

d)

5 4 2 9

2 4 6

8 3 3 2

x y z

x y z

x y z

− + =

+ − = − − − =

e)

3 2 4

2 3

4 3 7

x y z

x y z

x y z

+ + =

− + = + + =

Challenge Problem

For the systems in exercises 1.1 and 1.2, that are dependent,

1) Describe solutions in parametric form.

2) Find a particular solution.

(Answers may vary).

7

Session 1A: Determinants and Cramer’s Rule

Session1A

Determinants and Cramer’s Rule

In the previous session, we solved systems of three linear equations by the elimination method. This method requires some specific operations performed on the equations of given systems. In this session, we consider formulas that allow us to explicitly calculate solutions by directly substituting the coefficients of the equations into these formulas, instead of manipulating the equations. Such formulas are called the Cramer’s rule named after Gabriel Cramer (1704 – 1752), a Swiss mathematician. Cramer’s rule is not efficient for systems with many equations, and it is not used in practice. However, it is easy enough to use for systems with two and three equations that we consider here. Cramer’s rule is especially convenient when the coefficients of given system are integers, but the solutions are fractions.

Case of a system with two equations We derive Cramer’s rule for the system

⎩⎨⎧

=+

=+

feydxcbyax

First, we solve this system by elimination method. Let’s eliminate the variable y by multiplying the first equation by e, the second equation by –b, and adding the resulting equations:

⎩⎨⎧

=+

=+

feydxcbyax

⇒ ⎩⎨⎧

−=−−

=+

bfbeybdxcebeyaex

Add the equations on the right, and solve for x: aex – bdx = ce – bf ⇒ (ae – bd)x = ce – bf. So, if 0ae bd− ≠ ,

bdaefbcex

−

−= .

In similar way we can find y by eliminating x:

⎩⎨⎧

=+

=+

feydxcbyax

⇒ ⎩⎨⎧

=+

−=−−

afaeyadxcdbdyadx

Add the equations on the right, and solve for y: – bdy + aey = –cd + af ⇒ (ae – bd)y = af – cd. So, if 0ae bd− ≠ ,

bdaecdafy

−

−= .

-b e

– da

8


We have come up with the following general formulas for the solutions of a system of two linear equations with two variables:

bdaefbcex

−

−= ,

bdaecdafy

−

−= , (when 0ae bd− ≠ ).

Notice that the denominators of both fractions are the same, and the structure of the numerators looks similar to the denominators. Cramer’s rule represents these formulas in terms of a special number that is called the determinant. The determinant is defined by four numbers, say k, l, m, and n. Here is the notation and the definition of the determinant:

mlknnmlk

−= .

We call this a 22× determinant. As you can see, to calculate the determinant, we take the product along the main diagonal (from left top corner to right bottom corner, so we multiply k by n) minus the product along the minor diagonal (from left bottom corner to right top corner, so we multiply m by l). If you return to the formulas for x and y, you may notice that their numerators and denominators can be written in terms of determinants. We thus derived the following rule, which is called the Cramer’s rule: To solve the system

⎩⎨⎧

=+

=+

feydxcbyax

we proceed in the following three steps:

1) Calculate the determinant D which is called the determinant of the system:

bdaeedba

D −== .

Notice that the coefficients c and f from the right side of the system are not used in this determinant. It consists of the coefficients for x and y only.

2) Calculate another two determinants, xD and yD :

bfceefbc

Dx −== , cdaffdca

Dy −== .

Notice that the determinant xD is obtained from D by replacing its first column with the column of the coefficients c and f. Similarly, the determinant yD is obtained from D by replacing its second column with the column of “free” coefficients.

3) If 0D ≠ , calculate the solution of the system by the formulas

9


DDx x= ,

DD

y y= .

Note. As you see, the denominator in these fractions is the determinant D. Therefore, these formulas make sense only if 0≠D . If D = 0, then the system does not have a unique solution. Instead, it either does not have solutions at all (the system is inconsistent), or it has infinitely many solutions (the system is dependent). To detect which case we have, we should check xD (or yD ) for zero. If 0≠xD , then there are no

solutions. If 0=xD , then the system has an infinite number of solutions. (If D = 0, then both xD and yD are equal or not equal to zero simultaneously, see exercise 2.7).

Example 1A.1. Solve the following system using the Cramer’s rule.

⎩⎨⎧

=+

=−

735427

yxyx

.

Solution.

1) Calculate the determinant D of the system:

311021)2(5373527

=+=−⋅−⋅=−

=D .

2) Calculate the determinants xD and yD :

261412)2(7343724

=+=−⋅−⋅=−

=xD ,

.29204945777547

=−=⋅−⋅==yD

3) Write the solution of the system:

3126

==DDx x ,

3129

==DD

y y .

Final answer: 3126

=x ,3129

=y , or, as a pair,26 29,31 31

⎛ ⎞⎜ ⎟⎝ ⎠

.

Case of a system with three equations

We will describe the Cramer’s rule for the system

⎪⎩

⎪⎨

⎧

=++

=++

=++

3333

2222

1111

dzcybxadzcybxadzcybxa

.

10


Similar to systems with two equations, the solutions of this system can also be represented in terms of determinants as ratios of determinants xD , yD , and zDcorresponding to the variables x, y, and z, to the common determinant D of the system. We now describe how to find these determinants. We will not derive these formulas in this text, but just provide the final result. The determinant D of the above system is denoted by

333

222

111

cbacbacba

D = .

This is a 33× determinant constructed from the coefficients of the system. There are several methods to calculate it. We only consider here one method: a direct calculation.

Direct calculation method. Here is the formula for the determinant D:

132321321 cbaacbcbaD ++=

132321321 abccababc −−− .

This formula seems difficult to memorize. Notice that it contains six terms: three terms with the plus sign, and another three with the minus sign. Here is one of the possible ways to remember this formula. Extend (double) the determinant D to the following table:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

333333

222222

111111

cbacbacbacbacbacba

.

To get the three terms of the determinant with the plus sign, calculate products along the main diagonal 321 ,, cba , and two parallel diagonals 321 ,, acb and 321 ,, bac .

To get the three terms with the minus sign, calculate products along the minor diagonal 1 2 3, ,c b a , and two parallel diagonals 1 2 3, ,a c b and 1 2 3, ,b a c . Symbolically, we multiply

coefficients along the lines

* * * * * ** * * * * ** * * * * *

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

* * * * * ** * * * * ** * * * * *

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Terms with the plus sign Terms with the plus sign

Note. The last column of the above table is not used, so it is not necessary to write it.

11


Example 1A.2. Calculate the following determinant by direct calculation

302423265

−

−−

=D .

Solution. Construct the extended table (we dropped the last column)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−−

023022342365265

We have 03)2(2)4()6(325 ⋅⋅−+⋅−⋅−+⋅⋅=D

1405484830)6(335)4(0)2(22 =+++=−⋅⋅−⋅−⋅−−⋅⋅− .

Now, we are ready to describe the Cramer’s rule for the system

⎪⎩

⎪⎨

⎧

=++

=++

=++

3333

2222

1111

dzcybxadzcybxadzcybxa

.

To solve the above system we proceed in the following three steps:

1) Calculate the determinant D of this system:

333

222

111

cbacbacba

D = .

2) Calculate the three other determinants, xD , yD and zD that correspond to variables x, y, and z. These determinants are constructed by replacing corresponding columns of the determinant D with the column from the right side of the system:

333

222

111

cbdcbdcbd

Dx = ,

333

222

111

cdacdacda

Dy = ,

333

222

111

dbadbadba

D = .

3) If 0D ≠ , calculate the solution of the system via the formulas

, ,yx zDD Dx y z

D D D= = = .

Note. Similar to the case of the system with two variables, either there is no solution, or there are an infinite number of solutions if D = 0.

12


Example 1A.3. Solve the following system using the Cramer’s rule.

⎪⎩

⎪⎨

⎧

=+

−=−+

=−−

53284237265

zxzyxzyx

Solution.

1) The determinant D of the system is302423265

−

−−

=D .

This is exactly the same determinant as in example 13.2, so D = 140.

2) Calculate the determinants xD , yD and zD :

305428267

−−

−−

=xD , 352483275

−−

−

=yD , 502823765

−

−

=zD .

For determinate xD , construct the extended table

7 6 2 7 68 2 4 8 25 0 3 5 0

− − −⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥⎣ ⎦

.

7 2 3 ( 6) ( 4) 5 ( 2) ( 8) 05 2 ( 2) 0 ( 4) 7 3 ( 8) ( 6) 38

xD = ⋅ ⋅ + − ⋅ − ⋅ + − ⋅ − ⋅

− ⋅ ⋅ − − ⋅ − ⋅ − ⋅ − ⋅ − =

For determinate yD , construct the extended table

5 7 2 5 73 8 4 3 82 5 3 2 5

−⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥⎣ ⎦

.

5 ( 8) 3 7 ( 4) 2 ( 2) 3 5

2 ( 8) ( 2) 5 ( 4) 5 3 3 7 201yD = ⋅ − ⋅ + ⋅ − ⋅ + − ⋅ ⋅

− ⋅ − ⋅ − − ⋅ − ⋅ − ⋅ ⋅ = −

For determinate zD , construct the extended table

5 6 7 5 63 2 8 3 22 0 5 2 0

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

.

13


5 2 5 ( 6) ( 8) 2 7 3 02 2 7 0 ( 8) 5 5 3 ( 6) 208

zD = ⋅ ⋅ + − ⋅ − ⋅ + ⋅ ⋅

− ⋅ ⋅ − ⋅ − ⋅ − ⋅ ⋅ − =

3) Calculate the solutions x, y, and z of the system

38 19140 70

xDxD

= = = ,140201

−==DD

y y , 3552

140208

===DDz z .

Final answer: 3552,

140201,

7019

=−== zyx , or, as a triple, 19 201 52, ,70 140 35⎛ ⎞−⎜ ⎟⎝ ⎠

.

14


Exercises 1A

In exercises 1A.1 and 1A.2, solve the systems of equations using Cramer’s rule.

1A.1. 3 5 47 6 8x yx y− + =⎧⎨

− =⎩1A.2.

2 3 75 4 9x yx y− =⎧

⎨+ =⎩

In exercises 1A.3 and 1A.4, calculate the determinants.

1A.3. 3 2 15 4 32 1 7

D−

= − − 1A.4. 1 6 77 5 34 3 2

D−

= −

−

In exercises 1A.5 and 1A.6, use the results of exercises 1A.3 and 1A.4 respectively to solve the systems of equations using Cramer’s rule.

1A.5.

3 2 15 4 3 02 7 2

x y zx y zx y z

− + =⎧⎪− + − =⎨⎪ + + =⎩

1A.6.

6 7 27 5 3 14 3 2 0

x y zx y zx y z

+ − =⎧⎪

− + = −⎨⎪ − + =⎩

Challenge Problem

1A.7. As we described for the 22× system ⎩⎨⎧

=+

=+

feydxcbyax

, determinants D, xD and yD

are defined by the formulas

,D ae bd= − xD ce bf= − , and yD af cd= − .

Let D = 0 and 0xD = . Show that if b or e is not zero, then 0yD = . Give an example when both b and e are zeros, and this statement is wrong.

15

Session 2: Quadratic Equations: Factored Form

Session 2

Quadratic Equations: Factored Form

When we solve the linear equation ax + b = 0 with 0a , it always has a unique solution

a

bx −= . There are problems in which more complicated equations appear that may have

more than one solution.

Example 2.0. Suppose that you need to measure a piece of land in the shape of rectangle,

having a given area A and a given perimeter P. What are the sides of this rectangle?

Solution (equation only). Let’s denote the sides of the rectangle by variables x and y. Then Ayx = (area), and Pyx =+ 22 (perimeter). We can solve the last equation for y:

2y = P – 2x, and 2

2xPy

−= . If we substitute this expression for y into the first equation

Ayx = , we will get

( ) 2 222 2 2 2 2 2 0

2

P xx y x A x P x A xP x A x Px A

− = = − = − = − + = .

We come up with an equation that contains 2x . We read this expression as “x squared”. By

definition, 2x x x= . We will study expressions like this in details in session 3.

Informally, equations that contain 2x and are similar to the above, are called the quadratic

equations (as opposed to linear equations: ax + b = 0).

More precisely, an equation for the variable x is called the quadratic equation, if it can be

written in the form

0,02 =++ acbxax .

This form is called the standard form. Here a, b, and c are constant numbers which are

called the coefficients: a is called the leading, and c is called free (free of x) coefficient.

Note. If we omit the restriction 0a , the above equation will not necessarily be a

quadratic: when a = 0, the equation becomes linear 0bx c+ = . Therefore, when dealing

with quadratic equations, we will always assume that the leading coefficient 0a .

Coefficients b and c may be any real numbers, including 0. Also notice that in the standard

form the right side of the equation is equal to zero.

The quadratic equation can be given in various forms, but if necessary, it can always be

presented in a standard form.

Example 2.1. Write the following equations as quadratic equations in standard form.

Identify the coefficients a, b, and c.

1) (2x – 1)(x + 5) = 0.

2) 5)23( 2 =+x .

16


Solution. In both equations we just need to distribute, combine like terms, and bring all

terms from the right side to the left (if needed).

1) 5925102)5)(12( 22 −+=−−+=+− xxxxxxx .

We get the standard form

22 9 5 0,x x+ − = so, a = 2, b = 9, c = – 5.

2) 2 2 2 2(3 2) 5 9 12 4 5 9 12 4 5 0 9 12 1 0x x x x x x x+ = + + = + + − = + − = .

We get the standard form

29 12 1 0,x x+ − = so, a = 9, b = 12, c = – 1.

Notes.

1) In Example 2.1, 1), we say that the equation is written in the factored form.

2) In Example 2.1, 2), we say that the equation is written in the squared form. We

will discuss this form in session 9.

3) In solving Example 2.1, 2), we used the following formula (square of the sum

formula):

Another useful formula is the square of the difference formula:

4) We will also use the following formula which is called the difference of squares:

This formula tells us how to factor the difference of two squares. Try to memorize the

above three formulas. When solving a quadratic equation, it is not necessary to always represent it in the standard

form. In some cases, other forms may be preferable, as in Example 2.1: the factored form

or the squared form. These are the forms in which the quadratic equation can be solved

easier than in the standard form. Here we consider the factored form, and in session 9 – the

squared form.

Factored Form of the Quadratic Equation

In general, the factored form of the quadratic equation looks like the following

(mx + n)(px + q) = 0.

The method to solve this equation is based on the following simple observation: if the

product of two values A and B is zero, i.e. 0=BA , then at least one of them is zero:

2 2 ( )( )a b a b a b− = − +

( )2 2 22a b a ab b+ = + +

( )2 2 22a b a ab b− = − +

17


A = 0 and/or B = 0. Using this property (which is called the zero-product property), the

equation (mx + n)(px + q) = 0 can be split into two linear equations:

mx + n = 0 and px + q = 0,

which can be solved to get the solutions to the original equation.

Example 2.2. Solve the equation from Example 2.1, 1): (2x – 1)(x + 5) = 0.

Solution. Since this equation is written in factored form, using the zero-product property

it can immediately be split into two equations: 2x – 1 = 0 and x + 5 = 0. From the first

equation, 2

1=x , and from the second, x = – 5. So, the original equation has two solutions:

2

1=x and x = – 5.

Many quadratic equations (but not all, if we use only rational numbers) can be solved by

factoring. Using this method, we first represent the given equation in the factored form,

and then split it into two linear equations like in Example 2.2. We consider separately two

cases for quadratic equations: the case when the leading coefficient 1a = , and the case

when 1a .

Case: leading coefficient 1a = .

In this case the standard form of the equation is

02 =++ cbxx .

Such an equation (when the leading coefficient is 1) is called the reduced equation. To

factor, we need to represent the left side as a product of two linear expressions (two pairs

of parentheses): ( )( ) 0x p x q+ + = . Let’s distribute the left side and combine like terms:

2 2( )( ) 0 ( ) 0x p x q x px qx pq x p q x pq+ + = + + + = + + + = .

If we compare the last equation with the original 02 =++ cbxx , we conclude that

p + q = b, and cqp = . This conclusion gives us an idea: to factor, we need to find two

numbers p and q such that their sum is the middle coefficient b and the product is the last

coefficient c.

Technically, to factor we can start with a template (skeleton) for the equation:

(x + __)(x + __) = 0.

Then consider all possible ways to factor the last coefficient c, and select such factors

whose sum is b. Replace the blanks with these numbers.

Example 2.3. Solve the quadratic equation 0652 =++ xx by factoring.

Solution. Start with the template (x + __)(x + __) = 0. For the last coefficient 6, there are two ways to factor: 6 2 3= and 6 1 6= (we ignore here negative numbers). We select

factors 2 and 3 since their sum is the middle coefficient 5. Replacing blanks with these

numbers, we get the factored form

18


(x + 2)(x + 3) = 0.

Now, using the zero-product property, split it into two equations: 2 0x + = and 3 0x + = .

Solve each and get two solutions:

x = – 2 and x = – 3.

Note. Keep in mind that in the factored form, the right side of given equation must be zero.

For example, the equation (x – 1)(x + 2) = 4 is NOT written in factored form and cannot

be split immediately into two linear equations.

Example 2.4. Solve the above equation (x – 1)(x + 2) = 4 by factoring.

Solution. To write this equation in factored form, we first represent it in standard form by

distributing the left side and combining like terms

4222 =−−+ xxx , or 062 =−+ xx .

Now to factor, we write the template (x + __)(x + __) = 0, and find two numbers such that

the product is – 6 and the sum is 1 (which is the coefficient for x). We can find 3 and – 2.

Substitute these numbers for the blanks and get the factored form (x + 3)(x – 2) = 0. Solving

the equations x + 3 = 0 and x – 2 = 0, we get two solutions:

x = – 3 and x = 2.

Example 2.5. Solve the quadratic equation 072 =+ xx by factoring.

Solution. Here the coefficient c = 0. We can factor the left side just by taking x out of

parentheses: x(x + 7) = 0. From here, x = 0 and 7 0 7x x+ = = − . The final answer is:

x = 0 and x = – 7.

Example 2.6. Solve the quadratic equation 23 48 0x − = by factoring.

Solution. Here the leading coefficient is not 1, it is 3. Notice, however, that both

coefficients 3 and –48 divisible by 3, and we can factor this 3 out: ( )23 16 0x − = . From

here we conclude that the expression inside the parentheses must be zero, so we may drop

the factor 3 and get an equation with the leading coefficient 1: 2 16 0x − = . Another way to

get this equation is just to divide all terms of the original equation by 3. Now, to factor the

left side of this equation, we may recognize it as the difference of two squares 2 2a b− with

2 2a x= and 2 16b = . Using the formula ( )( )2 2a b a b a b− = − + , we write the equation

2 16 0x − = in the factored form ( )( )4 4 0x x− + = . Now it remains to solve the two

equations 4 0x − = and 4 0x + = , and get the final answer: x = 4 and x = – 4. The final

answer can also be written as 4x = , meaning that we have combined both roots in one

formula.

Case: leading coefficient 1a .

We will show a method how to reduce the quadratic equation 02 =++ cbxax when 1a

to the case 1a = in such a way that if coefficients a, b and c are integers, then coefficients

19


of the reduced equation are also integers. In some cases, it is easier to solve the reduced

equation. The method is based on the following

Proposition. Let r be a root of the equation 2 0x bx ac+ + = . Then r

a is a root of the

equation 2 0ax bx c+ + = .

Proof. By definition of the root r, 2 0r br ac+ + = . Divide both sides by a:

22 2

20 0 0.

r br ac r r r ra b c a b c

a a a a a a a

+ + = + + = + + =

The last equality means that r

a is a root of the equation 2 0ax bx c+ + = . ∎

This proposition allows us to solve the equation 2 0ax bx c+ + = in the following three

steps:

1) Construct (temporary) a new reduced equation 02 =++ acbxx . In words: take away

coefficient a from 2x and multiply it by c.

2) Solve the above reduced equation. Let’s its roots be r and s.

3) Divide both r and s by a to get the roots of the original equation 2 0ax bx c+ + = . The

roots are r

a and

s

a.

Example 2.7. Solve the equation 0456 2 =−+ xx .

Solution. We use the above method.

1) Construct the reduced equation 2 5 6 4 0x x+ − = or 02452 =−+ xx .

2) Solve the reduced equation by factoring. To factor, we need to find two numbers such

that the product is –24 and the sum is 5. These numbers are 8 and –3. The above reduced

equation is factored as: (x + 8)(x – 3) = 0. Its roots are –8 and 3.

3) Divide both –8 and 3 by the leading coefficient 6 of the original equation to get its

roots. The roots are 8 4

6 3

−= − and

3 1

6 2= .

20


Exercises 2

In exercises 2.1 and 2.2, write the given equations as quadratic equations in standard form.

Identify the coefficients a, b, and c.

2.1 a) (3 7)( 2) 0x x+ − =

b) 2(4 3) 6x − =

2.2. a) (4 3)(2 1) 0x x− + =

b) 2(5 2) 3x + =

In exercises 2.3 and 2.4, solve the given equations.

2.3. a) (3 7)( 2) 0x x+ − =

b) 2(6 5) 0x − =

2.4. a) (4 3)(2 1) 0x x− + =

b) 2(7 4) 0x + =

In exercises 2.5 and 2.6, solve the given equations by factoring.

2.5. a) 23 5 0x x+ =

b) 22 32 0x − =

c) 2 12 0x x+ − =

d) 2 10 25 0x x+ + =

e) ( 4)( 3) 8x x+ − =

2.6. a) 26 7 0x x− =

b) 25 45 0x − =

c) 2 2 15 0x x− − =

d) 2 12 36 0x x+ + =

e) ( 3)( 5) 9x x+ − =


2.7. a) 28 2 3 0x x− − =

b) 25 12 9x x+ =

c) 29 2 9x x+ =

2.8. a) 26 13 5 0x x+ − =

b) 27 3 4x x+ =

c) 212 2 5x x− =

Challenge Problems

In problems 2.9 and 2.10, a and b are two positive numbers.

2.9. Which expression is bigger: 𝑎2 + 𝑏2 or (𝑎 + 𝑏)2?

2.10. Prove that 𝑎2+𝑏2

2 ≥ 𝑎𝑏.

2.11. Can you calculate the following expression mentally without using a calculator?

123452 − 123442

2.12. Construct a quadratic equation with

a) roots 2 and –3.

b) one root 4.

How many quadratic equations with the above roots can you construct?

21

Session 3: Integer Exponents

Session 3

Integer Exponents

Exponential Expressions with Positive Integer Powers

Let’s recall the well-known notation of multiplication. Everybody knows that 3 4 12 = .

But what does exactly multiplication mean? Why is the result 12? We got this result by

adding number 3 to itself 4 times:

3 4 3 3 3 3 12 = + + + = .

Multiplication means repetition with addition. It allows us to write the summation of a

number with itself in a short, compact form.

There are cases when we need repetition with multiplication. In other words, we want to

multiply a number by itself several times. For example, consider the product 3 3 3 3 .

It would be a good idea to invent a special notation, similar to the notation for

multiplication, that allows to write such a product in a short form, using the number 3

only one time (which tells us the number we want to multiply by itself) and number 4

(which tells us how many time to multiply). We cannot use the notation 43 because it

is already taken for multiplication to express repetition with summation. The following

notation was invented to express repetition with multiplication: 43 . This expression is

called the exponential expression. So, by definition

43 3 3 3 3= .

Note. In some computer languages and calculators, to keep both numbers on one line, the

notation 3^4 is used.

In similar way, we can define the exponential expression in general form:

Definition. For arbitrary number a and arbitrary positive integer n the exponential

expression na is defined by the formula:

...na a a a= (multiply n times)

The number a is called the base, and n is called the exponent or the power of the

expression. We can say that we raise a to the power n. In particular, 1a a= (we “repeat”

number a one time). Also, 1 1n = for any n. For two special cases, when power n = 2 and

n = 3, we also say that 2a is “a-squared”, and

3a is “a-cubed” respectively. The reason

for that is 2a represents the area of a square, and

3a represents the volume of a cube

with sides a.

The notation of exponents is useful in many situations, in particular, when we work with

very big numbers (for example, with distances between planets). Below we show a way

in which exponential expressions can also be used for very small numbers (such as, for

example, distances inside molecules or atoms).

Let’s consider examples and study some properties of exponents.

22


Example 3.1. Some people believe that one kilobyte (KB) of computer memory is equal

to 1000 bytes (B). However, 1 KB = 1024 B, not 1000 B. The reason is that the number

1024 is a power of 2 but the number 1000 is not. Express number 1024 in exponential

form with the base of 2.

Solution. Let’s construct a table of exponential expressions with the base of 2 starting

with the power of 5:

Power 5 6 7 8 9 10

Expression 25 = 32 26 = 64 27 = 128 28 = 256 29 = 512 210 = 1024

Therefore, 1024 = 210.

Example 3.2. It is known that the distance from our Earth to the Sun is about

150,000,000 km (150 million kilometers). Represent this distance in a short form using

exponents.

Solution. We can write this number in the form 150,000,000 1.5 100,000,000= . The

number 100,000,000 contains 8 zeros and can be written as 8100,000,000 10= .

Therefore, 8150,000,000 1.5 10= .

Note. Representation of big numbers such as in the example 3.2 in exponential form with

the base of 10 is widely used in science. This form is called scientific notation.

In general, we say that a positive number n is in scientific notation, if it is written as

product of two parts:

1) A number between 1 and 10 (1 is included, but 10 is not).

2) Power of 10.

Example 3.3. Consider three numbers: 815.3 10 ,

60.15 10 , and 42.73 10 . Are these

numbers in scientific notation?

Solution. It may seem that all three numbers are in scientific notation. However, it is not

true. The first number 815.3 10 is not in scientific notation, because its first part, the

number 15.3, is greater than 10, so it does not satisfy condition 1). The second number 60.15 10 is also not in scientific notation, since its first part, the number 0.15, is less

than 1, so again it does not satisfy condition 1). The third number 42.73 10 is in

scientific notation: its first part is 2.73, which is between 1 and 10.

Now, how about very small numbers? Consider, for example, the diameter of DNA helix.

It is known that this diameter is about 0.0000002 cm. Is it possible somehow to represent

this number also in a short form using exponents? The answer is yes. We will solve this

problem in example 3.5 below. To come up with the idea how to do this we need to l earn

more about exponents. Let’s start with some basic properties.

23


Basic Properties of Exponents

We will not give proofs here since proofs are very simple and follow directly from the

definition of exponents (if you wish you can try to prove yourself). We will just illustrate

the properties with some examples. Consider how we can combine the product of

2a and 3a into one expression:

( )( )2 3 5a a a a a a a a = = . As you can see, to combine, we add powers 2 and 3, but

not multiply them. Similar property holds for the following general rule:

Product Rule. For any number a, and any positive integers n and m,

n m n ma a a + = .

Note that both exponential expressions in this formula have the same base a. This

restriction is very important. If, for example, you need to multiply 4 53 2 , there is no

simple rule to represent the answer as a single expression. Also notice how product rule

works: to multiply expressions with the same base, we add powers. A possible mistake

here is multiplying powers instead of adding them.

Another rule is how to raise exponential expressions into a power. Consider the

expression ( ) ( ) ( ) ( ) ( )2

3 3 3 6a a a a a a a a a a= = = . This time, contrary to product

rule, we multiply powers 3 and 2. Here is the general rule:

Power Rule. For any number a, and any positive integers n and m,

( )m

n n ma a = .

Now let’s consider an example of dividing the exponential expressions:

66 2 4

2

a a a a a a aa a a a a a a

a a a

= = = =

.

To divide exponential expressions with the same base, we subtract powers (we do not

divide them). The general rule is this:

Quotient Rule. For any nonzero number a, and any positive integers n and m, such that n m ,

nn m

m

aa

a

−= .

Note that in the above formula, the power in the numerator is greater than that in the

denominator. But what if we need to divide expressions when the power of numerator is less than the power of denominator: n m ? One possible way is just to reduce this

fraction by dividing numerator and denominator by na . Consider the example

2

6 4

1 1a a a

a a a a a a a a a a a a

= = =

.

In general, if n m , then

24


1n

n m

m m n

aa a

a a − = = .

In the next example, we will see how to raise the product of two numbers to a power:

( ) ( )( )( ) ( )( )3 3 3ab ab ab ab aaa bbb a b= = = .

As you can see, we raise to the power 3 both numbers a and b. Here is the general rule.

Power of Product Rule. For any two numbers a and b, and any positive integer n,

( )n n nab a b= .

In similar way, if we have the quotient of numbers a and b, then to raise it to a power, we

raise to that power both a and b, as in the following example:

2 2

2

a a a aa a

b b b bb b

= = =

.

In general, the following rule is true.

Power of Quotient Rule. For any number a, any nonzero b, and any positive integer n,

n n

n

a a

b b

=

.

Negative Integer Exponents

Let’s reconsider the Quotient Rule when the power n of the numerator is less than the

power m of the denominator: 1n

m m n

a

a a −= , n m . It would be a good idea to somehow

write this in the same exponential form as in the case when n m :

nn m

m

aa

a

−= . In doing

this, we come up to the expressions with negative powers! For example, we can write

22 6 4

6 4

1aa a

a a

− −= = = , or 4

4

1a

a

− = .

You might say that negative exponent (expression with negative power) does not make

any sense. Indeed, by the initial definition of exponential expressions, its power tells us

how many times the base should be multiplied by itself. How can we multiply anything

“negative number of times”? Of course, we cannot. However, there is a way to give a

sense to the above formula with negative exponent.

The idea is to define expressions with negative exponents similar to the above example

for 4a−.

Definition. For any nonzero number a, and for any positive integer n, we define na−

as

25


1n

na

a

− = .

So, expressions with negative exponents are reciprocals to expressions with positive

exponents. In particular,

n na b

b a

−

=

.

Now, let’s try to invent the definition of 0a (exponent is zero).

We can use similar approach as for negative exponents, using Quotient Rule for m = n.

We will have 0.

nn n

n

aa a

a

−= = Because 1n

n

a

a= , we get

0 1.a = We come up with the

Definition. For any nonzero number a, 0 1.a =

Example 3.4. Calculate

a) 010 , b)

110−, c)

210−, d) 10 n−

.

Solution. By definition, we have

a) 010 1= , b)

1 110 0.1

10

− = = , c) 2

2

1 110 0.01

10 100

− = = = ,

d) 1

10 0.0...0110

n

n

− = = (n – 1 zeros after the decimal point).

Example 3.5. The diameter of DNA helix is about 0.0000002 cm. Represent this number

in exponential form.

Solution. 70.0000002 2 0.0000001 2 10−= = .

This example shows that expressions with negative exponents are useful for the

representation of small numbers in a compact form. The above representation, as for big

numbers, is also called scientific notation. In conclusion, consider several examples. It can be shown that all the above properties of

expressions with positive exponents are also true for negative exponents. In all problems

below, the question is to simplify given expression and write the answer using positive

exponents only.

Example 3.6.

n

m

a

a− .

Solution. We can use the Quotient Rule: ( )

nn m n m

m

aa a

a

− − +

−= = .

Technically, we can get the same answer as if we would “physically” move ma−

from the

denominator up to the numerator, and change the negative sign of m to positive. Then we

can use the Product Rule:

nn m n m

m

aa a a

a

+

−= = . We will use similar method in the

following example.

26


Example 3.7.

n p

m q

a b

a b

−

−.

Solution. We can get rid of negative exponent for ma−

in denominator, and for pb−

in

numerator, by moving these expressions into the opposite part of the fraction: move ma−

up to the numerator and move pb−

down to the denominator. Then apply the Product

Rule. We will get

n p m n m n

m q p q p q

a b a a a

a b b b b

− +

− += = .

Example 3.8. ( )( )5 3 1ax y bxy− −.

Solution. We can use Product Rule to combine 5x−

and x (note that x can be written as 1x ), and combine

3y and 1y−:

( )( )5 3 1 5 1 3 1 4 2ax y bxy abx y abx y− − − + − −= = .

Now, to get rid of negative exponent 4x−

, similar to Example 3.7, move 4x−

down:

24 2

4

abyabx y

x

− = .

Example 3.9.

23 8

6 4

45

18

u v

u v

−−

− −

.

Solution. It is possible to simplify this expression in different ways. As а first step, let’s

get rid of negative power 2− by applying the definition of negative exponent: take

reciprocal of the fraction inside parentheses:

2 23 8 6 4

6 4 3 8

45 18

18 45

u v u v

u v u v

−− − −

− − −

=

.

Next, we simplify fraction inside parentheses by reducing coefficients 18 and 45 by 9,

and moving both exponents 6u−

and 4v−

down (it is not needed to move 3u−

up). Then

we use Product Rule:

2 2 2 26 4

3 8 3 6 8 4 3 6 8 4 3 12

18 2 2 2

45 5 5 5

u v

u v u u v v u v u v

− −

− − − + +

= = =

.

Finally, we use Power of Quotient and Power Rules:

( ) ( )

2 2

2 23 12 3 2 12 2 6 242 3 12

2 2 4 4

5 25 255u v u v u vu v

= = =

.

27


Exercises 3

In exercises 3.1 –3.4, write the given numbers in scientific notation.

3.1. The Earth's circumference

at the equator is

approximately 25,000 mi.

3.2. Mount Everest (on the

border of Nepal and

China) is the highest

place on Earth above sea

level, at about 29,000 ft.

3.3. A tiny space inside a

computer chip has been

measured to be

0.000256 cm wide.

3.4. A tiny space inside a

computer chip has been

measured to be 0.000014 cm

long.

In exercises 3.5 – 3.8, write the given numbers as ordinary numbers.

3.5. The length of a bacterium is

about 54 10− in.

3.6. The weight of a flea is

about 28.75 10− g.

3.7. The diameter of the Moon is

63.475 10 m.

3.8. The speed of sound in dry

air at is 31.236 10 km/h.

In Exercises 3.9 and 3.10, the numbers are not written in scientific notation (why?).

Write these numbers in scientific notation.

3.9. a) 434.7 10−

b) 30.25 10

3.10 a) 643.8 10−

b) 50.36 10

In Exercises 3.11 and 3.12, calculate without calculator and write the answer without

using of exponents.

3.11. a) 02 3

b) ( )0

2 3

c) 24−

d) 24−−

e) ( )2

4−

−

3.12. a) 03 4

b) ( )0

3 4

c) 25−

d) 25−−

e) ( )2

5−

−

28


In Exercises 3.13 – 3.24, simplify and write the answer using positive exponents only

(assume that all letters represent positive numbers).

3.13. a) 5 7a a−

b) 3 6c c− −

c) ( )2

3a−

d)

2m

n

−

−

3.14. a) 6 3c c−

b) 2 4n n− −

c) ( )3

4d −

d)

3a

b

−

−

3.15. a)

8

4

p

p

b)

8

4

p

p

−

c)

8

4

p

p−

d)

8

4

p

p

−

−

e)

4

8

p

p

f)

4

8

p

p−

g)

4

8

p

p

−

h)

4

8

p

p

−

−

3.16. a)

9

3

r

r

b)

9

3

r

r−

c)

9

3

r

r

−

d)

9

3

r

r

−

−

e)

3

9

r

r

f)

3

9

r

r

−

g)

3

9

r

r−

h)

3

9

r

r

−

−

3.17.

a b

c d

x y

x y

−

−

3.18.

w x

y z

p q

p q

−

−

3.19 ( )( )4 2 5 3ma b na b− − 3.20. ( )( )5 3 7 6xu v yu v− −

3.21.

26 12

2 3

35

42

r s

r s

−−

−

−

3.22.

315 9

5 3

12

30

x y

x y

−−

−

−

3.23.

38 14

2 7

24

16

p q

p q

−−

−

−

3.24.

212 8

2 4

25

15

m n

m n

−−

−

−

29

Session 4: Rational Expressions and Complex Fractions

Session 4

Rational Expressions and Complex Fractions

Recall that a rational number is a number that can be written as a fraction of integers

(having a numerator on the top and a denominator on the bottom). Usually, we write a

fraction in the form m

n, where m and n are two integers (m is the numerator, and n is the

denominator and 0n ). We treat a fraction as a ratio of its numerator to denominator, so,

we can write m

m nn= . We will always assume that the denominator n is not equal to

zero.

We consider here rational expressions. These are also fractions. However, their

numerators and denominators are not necessarily numbers. They are expressions that are

called polynomials. A polynomial is an expression that can be written in a form that

contains a variable, say x, together with the operations of addition, subtraction, and

multiplication of x by numbers and by itself. When x is multiplied by itself, (like

,x x x x x ,…), we usually write this as an exponential expression (like 2 3, ,...x x ).

Here are some examples of polynomials:

3 21

3, 5 3 2, 3 7 4.2

x x x x x− − + − +

The last polynomial is called a quadratic trinomial (quadratic, because the highest power

of x is 2, and trinomial, since it contains three terms). The expression 𝑥(3𝑥2 + 𝑥) is also

a polynomial, because after distribution it can be written as 3 23x x+ . On the contrary,

13

x+ is not a polynomial since it contains the variable x in the denominator (so, division

by the variable) and cannot be reduced to a polynomial.

Definition. A rational expression is a ratio of two polynomials (or a fraction whose

numerator and denominator are polynomials). Here are several examples of rational

expressions:

2

2 3 2

5 3 2 1 3 3 2, , , .

2 1 1 5

x x x

x x x x x

− + +−

− − −

Below we consider examples on how to add and subtract rational expressions. Mostly,

these operations can be performed in a manner similar to rational numbers (ordinary

fractions). Where possible, we will point out the similarity between rational numbers and

rational expressions.

30


Simplification of Rational Expressions

When adding or subtracting, we will also simplify (if possible) resulting expressions. In

order to simplify, we factor the numerator and denominator, and reduce (cancel out) a

common factor, as we were doing with rational numbers (ordinary fractions). Let’s see

some examples of simplification.

Example 4.1. Simplify 210 15

20

x x

x

−.

Solution. A possible mistake here is to cancel x from 15x and 20x and not from 210x , and

as a result one arrives at a wrong answer 2(10 15) / 20x − . As we mentioned above, the

correct way is to factor the numerator before reducing. We can factor 5x from the numerator

and then reduce: 210 15 5 (2 3) 5 (2 3) 2 3

20 20 5 4 4

x x x x x x x

x x x

− − − −= = =

.

Example 4.2. Simplify 216 12

6 10

x x

x

+

−.

Solution. Here we can factor out both the numerator and denominator and then reduce

(divide the numerator and denominator) by 2:

216 12 4 (4 3) 2 (4 3)

6 10 2(3 5) 3 5

x x x x x x

x x x

+ + += =

− − −.

Example 4.3. Simplify 2

2

5 30 40

3 3 18

x x

x x

+ +

− −.

Solution. As in the previous problem, we start with factoring the numerator and the

denominator. We can factor them in two steps. First, factor out 5 from the numerator and

3 from the denominator:

2 2

2 2

5 30 40 5( 6 8)

3 3 18 3( 6)

x x x x

x x x x

+ + + +=

− − − −.

The second step is to factor the quadratic trinomials in the parentheses:

2 6 8 ( 2)( 4)x x x x x+ + = + + and 2 6 ( 2)( 3)x x x x− − = + − .

We can complete the factorization and then reduce (cancel) the common factor x+ 2 :

5(x2 +6x+8)

3(x2 - x -6)=

5(x+ 2)(x+ 4)

3(x+ 2)(x-3)=

5(x+ 4)

3(x-3).

31


Addition and Subtraction of Rational Expressions

Example 4.4. Add 5 3

2 1 2 1

x

x x+

− −.

Solution. Recall that it is very easy to add (or subtract) numerical fractions if they have the

same denominator: just add (or subtract) numerators and keep (do not add or subtract) their

common denominator. For example, 2 3 5 5 7 2

, .7 7 7 9 9 9+ = − = −

The same rule applies to rational expressions. For given example,

5 3 5 3

2 1 2 1 2 1

x x

x x x

++ =

− − −.

Example 4.5. Subtract 3 4 2

8 6

a a+ −− .

Solution. This time the denominators are different. Recall how we would subtract rational

numbers (fractions), let’s say 7 5

8 6− .

To subtract, we replace these fractions with equivalent ones having the same denominator,

which is called the LCD (Least Common Denominator). LCD is the smallest number that

is divisible by both denominators. Technically, we can subtract fractions in three steps.

1) Find the LCD. For given denominators 8 and 6, the LCD = 24. We put the LCD into the

denominator of the resulting fraction.

2) Find complements of each denominator to the LCD. A complement is the number such

that if we multiply it by the denominator, we get the LCD. To find complements, just divide

the LCD by each denominator. For the denominator 8, the complement is 3 ( 3824 = ),

and for the denominator 6, the complement is 4 ( 4624 = ).

3) Calculate the numerator of the resulting fraction: multiply numerator of each fraction by

the complement to its denominator and subtract the results. For the given fractions 7/8 and

5/6, multiply the numerator 7 by 3 (complement to the denominator 8), and the numerator

5 by 4 (complement to the denominator 6):

7 5 7 3 5 4 21 20 1

8 6 24 24 24

− −− = = = .

To subtract rational expressions, we do the same thing:

3 4 2 (3 4)3 ( 2)4 9 12 4 8 5 20

8 6 24 24 24

a a a a a a a+ − + − − + − + +− = = = .

Now consider an example when the denominators are also different and contain variables.

32


Example 4.6. Add 3 4

10 15x y+ .

Solution. As before, we first construct the LCD. The denominators contain both numbers

and letters. Therefore, the LCD consists of two parts: numerical part and letter part. For the

numbers 10 and 15, the numerical part of the LCD is 30. The letters x and y do not have

common factors. Therefore, the letter part of the LCD is their product xy. The entire LCD

is the product of numerical and letter parts:

LCD = 30xy.

Next, we find complements of each denominator to the LCD by dividing the LCD by

denominators.

For the denominator 10x, the complement is 30xy/10x = 3y.

For the denominator 15y, the complement is 30xy/15y = 2x.

Finally, we add the given fractions:

3 4 3 3 4 2 9 8

.10 15 30 30

y x y x

x y xy xy

+ ++ = =

Example 4.7. Combine 2

5 11 9

6 14a a− + .

Solution. To construct the LCD, similarly to the previous example, we construct separately

its numerical and letter parts.

For the numbers 6 and 14 in denominators, the numerical part of the LCD is 42.

For the letters 2a and a, the letter part of the LCD is 2a .

The entire LCD is the product of both parts: LCD = 242a .

Next, we find complements for each denominator to the LCD:

For 2a , the complement is 2 242 / 42a a = .

For 6a, the complement is 242 / 6 7a a a= .

For 14, the complement is 2 242 /14 3a a= . From here,

2 2

2 2 2

5 11 9 5 42 11 7 9 3 210 77 27.

6 14 42 42

a a a a

a a a a

− + − +− + = =

Note. In general, if the denominators contain exponential expressions with the same base

and different powers, include into the LCD the expression with the biggest power. Thus,

in the previous example, for the expressions 2a and a in the denominators we have

included into the LCD 2a .

If denominators do not contain common factors, include into the LCD both denominators.

Example 4.8. Subtract 4 2

5 3 3 5x x−

− −.

33


Solution. The denominators 5 3x − and 3 5x − do not have common factors, therefore, the

LCD is simply their product:

LCD (5 3)(3 5)x x= − − .

The denominators 5 3x − and 3 5x − are complemented to each other, therefore

4 2 4(3 5) 2(5 3) 12 20 10 6 2 14

5 3 3 5 (5 3)(3 5) (5 3)(3 5) (5 3)(3 5)

x x x x x

x x x x x x x x

− − − − − + −− = = =

− − − − − − − −.


6 6

y y

y y

− +−

− −.

Solution. Notice that the denominators of these fractions are “almost the same” (they differ

only in signs and the order of terms). We can make them exactly the same by using the

following connection between expressions a b− and b a− :

( )a b b a− = − − .

Therefore, 6 ( 6)y y− = − − and

5 5 5 5 5 5 2

6 6 6 ( 6) 6 6 6

y y y y y y y

y y y y y y y

− + − + − +− = − = + =

− − − − − − − −.

Example 4.10. Add 7

43x+

−.

Solution. We can treat the integer 4 as a fraction with the denominator 1: 4

4 .1

=

From here, the LCD of the denominators 3x − and 1 is 3x − , and these denominators are

complemented to each other. Therefore,

7 7 4 7 1 4( 3) 7 4 12 4 5

43 3 1 3 3 3

x x x

x x x x x

+ − + − −+ = + = = =

− − − − −.

Example 4.11. Add 2

5 3.

4 2 4

x

x x+

− −

Solution. At the first glance, it might seem that the denominators 2 4x − and 2 4x − do

not have common factors. However, they do! To see that, factor both. It is easy to factor

the denominator 2 4x − by factoring the number 2: 2 4 2( 2).x x− = − The denominator 2 4x − can be factored using the formula for the difference of two squares:

2 2 ( )( )a b a b a b− = − + .

From here, 2 4 ( 2)( 2).x x x− = − + Now, compare the denominators in the factored form:

2( 2)x − and ( 2)( 2).x x− +

34


We see the common factor 2x − . This is a part of the LCD. Also, we put the other factors

2 and 2x + into the LCD. Therefore, the entire LCD 2( 2)( 2)x x= − + . Next, we find the

complements for each denominator:

For 2 4 ( 2)( 2),x x x− = − + the complement is 2.

For 2 4 2( 2),x x− = − the complement is 2x + .

Finally, we add the fractions:

2

5 3 5 2 3( 2) 10 3 6 13 6

4 2 4 2( 2)( 2) 2( 2)( 2) 2( 2)( 2)

x x x x x x

x x x x x x x x

+ + + + ++ = = =

− − − + − + − +.


7 5

12 6x x x x−

− − + −.

Solution. Again, as a first step, we factor each denominator:

2 12 ( 3)( 4)x x x x− − = + − and 2 6 ( 3)( 2)x x x x+ − = + − .

Now, to construct the LCD, we multiply all factors from both denominators (taking the

common factor 3x + only one time):

LCD = (x + 3)(x – 4)(x – 2).

Next, using the LCD, we find the complements of each denominator.

For 2 12 ( 3)( 4)x x x x− − = + − , the complement is x – 2.

For 2 6 ( 3)( 2)x x x x+ − = + − , the complement is x – 4.

Finally, we subtract the given fractions:

2 2

7 5 7( 2) 5( 4)

12 6 ( 3)( 4)( 2)

x x

x x x x x x x

− − −− =

− − + − + − −

= 7 14 5 20 2 6

( 3)( 4)( 2) ( 3)( 4)( 2)

x x x

x x x x x x

− − + +=

+ − − + − −.

We can simplify the last expression even more by factoring the numerator 2x + 6 as

2( 3)x + and then canceling the factor x + 3 from the numerator and the denominator:

2 6 2( 3) 2

( 3)( 4)( 2) ( 3)( 4)( 2) ( 4)( 2)

x x

x x x x x x x x

+ += =

+ − − + − − − −.

Example 4.13. Combine 2

2

6

4

124

32 +

−−

+−− aaaa

.

Solution. Factor the first denominator:

35


)2)(6(1242 +−=−− aaaa .

The factors a – 6 and a + 2 are the denominators of the second and the third fractions, so

LCD = (a – 6)(a + 2).

Since the LCD coincides with the first denominator, the complement of this is 1, and we

need to find complements only for the second and third denominators.

For a – 6, the complement is a + 2.

For a + 2, the complement is a – 6.

We have

)2)(6(

)6(2)2(43

2

2

6

4

124

32 +−

−−++=

+−

−+

−− aa

aa

aaaa

= )2)(6(

232

)2)(6(

122843

+−

+=

+−

+−++

aa

a

aa

aa.

Complex Fractions Complex fractions are fractions that contain other fractions in their numerators and/or

denominators. We will call fractions inside a complex fraction inner ones. For example,

fractions a

b and

c

d are inner fractions in the complex fraction:

a

bc

d

We will consider

methods on how to simplify complex fractions.

Example 4.14. Simplify 2

7

1514

27

ab

ab

.

Solution. We can represent this complex fraction in the form of division of its numerator

by its denominator: abab 27

14

15

72 . Now we use the rule for division of fractions: multiply

the first fraction by the reciprocal of the second: 14

27

15

72

ab

ab . After simplification

(reducing) we get the final answer b10

9.

Note. Working with complex fractions, it is important not to confuse those that look

similar, but in reality are different. We need to carefully identify the position of the “main”

(longest) fraction line. This line shows the place where we divide numerator by

denominator. For example, compare the following complex fractions (which have different

“longest” lines):

36


As you see, these fractions are different.

In the examples below, to simplify complex fractions, we use two methods.

First method: find the LCD for all inner fractions contained in the numerator and the

denominator of the complex fraction, and then multiply the denominator the denominator

by the LCD.

Second method: simplify the numerator and the denominator of a complex fraction

separately, and then divide its numerator by the denominator.

Example 4.15. Simplify

12

7

15

46

5

3

2

+

−.

Solution.

First method. The denominators of the four inner fractions are 3, 6, 15, and 12. Their LCD

is 60. Multiply the numerator and denominator of the complex fraction by this LCD:

51

10

51

10

3516

5040

12

760

15

460

6

560

3

260

12

7

15

460

6

5

3

260

−=−

=+

−=

+

−

=

+

−

.

Second method. Simplify the numerator and denominator separately:

6

1

6

522

6

5

3

2−=

−=− ,

60

51

60

5744

12

7

15

4=

+=+ .

Divide numerator by denominator: 51

10

51

60

6

1

60

51

6

1−=−=− .


22

2

45

23

xxy

xyyx

−

+

.

Solution.

First method. The denominators of the four inner fractions are 22 ,, xyxyyx , and 2x .

Their LCD is 22 yx . Multiply the numerator and the denominator of the above complex

fraction by this LCD:

.b

ac

b

ca

c

ba ===

a

b

c

a

b

c.

1

bc

a

cb

ac

b

a===

37


2

2

22

2

22

22

2

22

22

22

2

22

45

23

45

23

45

23

yx

xyy

x

yx

xy

yx

xy

yx

yx

yx

xxyyx

xyyxyx

−

+=

−

+

=

−

+

.

Second method.

Add inner fractions in the numerator: yx

x

xyyx 22

2323 +=+ .

Subtract inner fractions in the denominator: 22

2

22

4545

yx

yx

xxy

−=− .

Divide numerator by denominator:

222

22

222

2

2 45

23

45

)23(

45

234523

yx

xyy

yx

yx

yx

yx

yx

x

yx

yx

yx

x

−

+=

−

+=

−

+=

−

+.


2

37

2

56

−+

−−

n

n .

Solution.

First method. The LCD of the inner denominators is n – 2. Multiply the numerator and

denominator of the complex fraction by n – 2:

117

176

3147

5126

3)2(7

5)2(6

2

37)2(

2

56)2(

2

37

2

56

−

−=

+−

−−=

+−

−−=

−+−

−−−

=

−+

−−

n

n

n

n

n

n

nn

nn

n

n.

Second method.

Combine the numerator: 2

176

2

5)2(6

2

56

−

−=

−

−−=

−−

n

n

n

n

n.

Combine the denominator: 2

117

2

3147

2

3)2(7

2

37

−

−=

−

+−=

−

+−=

−+

n

n

n

n

n

n

n.

Divide numerator by denominator:

117

176

117

2

2

176

2

117

2

176

−

−=

−

−

−

−=

−

−

−

−

n

n

n

n

n

n

n

n

n

n.

38


Exercises 4

In exercises 4.1 – 4.8, simplify the given expressions.

4.1. 218 12

15

y y

y

−

4.2.

224 16

14

z z

z

−

4.3. 215 25

10 30

z z

z

+

−

4.4.

212 18

9 15

y y

y

+

−

4.5. 2

2

4 8 60

6 6 72

x x

x x

+ −

+ −

4.6.

2

2

6 24 72

8 40 48

x x

x x

+ −

+ −

4.7. 2

2

2 6 56

7 49

x x

x x

− −

−

4.8.

2

2

3 3 36

4 16

x x

x x

− −

−

4.9. Add and simplify

6 4

3 2 3 2

x

x x+

+ +


10 2

5 1 5 1

x

x x+

+ +

4.11. Subtract

4 3 3

12 18

m m+ −−

4.12. Subtract

2 5 4

10 12

n n+ −−

4.13. Add

5 7

8 12x y+

4.14. Add

7 3

6 8x y+

4.15. Combine

2

4 5 7

15 3 6x x− +

4.16. Combine

2

5 7 3

12 6 8x x− +

4.17. Subtract

3 5

4 7 7 4x x−

− −

4.18. Subtract

6 7

3 4 4 3x x−

− −

4.19. Subtract

2 2

4 4

a a

a a

− +−

− −

4.20. Subtract

4 4

5 5

b b

b b

− +−

− −

39


4.21. Add

6

32x+

−

4.22. Add

8

24x+

−


2

2 3 1

5 25 25

x x

x x

− ++

+ −


2

4 2 3

16 4 16

x x

x x

+ −+

− +

4.25. Subtract and simplify

2 2

3 1

7 10 5 6x x x x−

+ + + +

4.26. Subtract and simplify

2 2

6 5

4 5 3 4x x x x−

+ − + −

4.27. Combine and simplify

2

5 3 3

2 8 4 2b b b b+ −

+ − + −

4.28. Combine and simplify

2

7 4 4

2 15 3 5c c c c− +

+ − − +

In exercises 4.29 – 4.44, simplify the complex fractions.

4.29. 3

2

5

1215

16

c d

cd

4.30. 6

3 2

18

56

25

m n

m n

4.31. 2

2

3 2

4 5x x

x x

+

−

4.32. 2

2

2 3

5 4x x

x x

−

+

4.33.

7

53

2 10

x

x

x

+

+

4.34.

5

74

3 21

x

x

x

−

−

4.35.

2 3

5 47 1

10 2

−

+

4.36.

5 7

6 83 2

4 3

+

−

40


4.37. 2

2 2

4 3

6 2

xy xy

x y y

+

−

4.38. 2

2 2

6 5

4 3

xy x y

x xy

−

+

4.39.

56

25

26

x

x

−

−

4.40.

712

2

73

8

y

y

−

−

4.41.

74

35

63

k

k

−−

+−

4.42.

85

46

74

m

m

+−

−−

4.43. 2

2

1 21

6 81

x x

x x

+ −

+ +

4.44. 2

2

2 31

5 61

x x

x x

+ −

+ +

Challenge Problems

4.45. Let A a

B b= . Prove that

( )( )2 2

A B A B

x ax x bx x a x b

−− =

− − − −.

4.46. Let 𝑑 = 𝑏𝑐 − 2𝑎 − 2𝑏. Prove that

𝑥 − 𝑎

𝑏𝑥 + 𝑏2+

𝑐𝑥 + 𝑑

𝑥2 − 𝑏2=𝑥 + 𝑎 + 𝑑

𝑏(𝑥 − 𝑏).

4.47. Let a, b and c be positive integers such that b and c are greater than 1. Which value

is bigger

or ?

41

Session 5: Rational Equations

Session 5

Rational Equations

In the previous session, we worked with rational expressions. In this session we will work

with rational (fractional) equations. Both sides of such equations are sums or differences

of rational expressions. Here is an example: 5 3 7

4 4 6

x x+ −= − . Below we solve this

equation by reducing it to an equation with no fractions. As with expressions, we will use

the LCD to do this. However, the main technical difference here is that in the expression

we must keep the LCD (and write it in the denominator of the answer), while in an equation

we can drop the LCD.

The reason why we can drop the denominator is this. When keeping the LCD (as we do

with expressions), both sides of the equation become fractions with the same denominator

(which is the LCD). If two fractions are equal and have the same denominator, then their

numerators are also equal, so we equate numerators and drop denominators. Here is a

simple example: 5

4 4

x= . Form here we conclude that x = 5. So, technically, we drop the

common denominator 4 and equate the numerators.

Example 5.1. Solve the equation 5 3 7

4 4 6

x x+ −= − .

Solution. The first step is the same as for expressions: find the LCD. For the denominators

4 and 6, the LCD = 12 (notice that the number 24 is also common denominator but not the

least). The second step is also the same: find complements for each denominator to the

LCD.

For the denominator 4, the complement is 3,

For the denominator 6, the complement is 2.

The third step is again the same: multiply each numerator by the corresponding

complement. But now, unlike what we did with expressions, we may drop all denominators! As a result, the original equation becomes an equation with no fractions:

3( 5) 3 3 2( 7)x x+ = − − .

Now it is easy to solve it:

3 3 5 3 3 2 2 7 3 15 9 2 14 3 15 23 2x x x x x x+ = − + + = − + + = − .

From this point we collect all terms with the variable x on the left side, and all other terms

(numbers) on the right side by using the moving method:

8

3 2 23 15 5 85

x x x x+ = − = = .

Next we consider equations in which denominators contain a variable. The technique here

is the same. The only additional (and very important) thing is that we need to check that

42


the final answer does not cause any denominator of the original equation to become zero.

If this happens, we must reject such a solution.

Example 5.2. Solve the equation 2 1 11

7 6 14x x+ = .

Solution. For the denominators 7x, 6 and 14x, the LCD = 42x. Next we find complements

for each denominator to the LCD:

For 7x, the complement is 6.

For 6, the complement is 7x.

For 14x, the complement is 3.

We multiply each numerator by its corresponding complement and drop the LCD. The

equation becomes free of fractions:

2 6 1 7 11 3x + = Now we solve this equation:

12 7 33 7 33 12 7 21 3x x x x+ = = − = = .

None of the denominators of the original equation is zero for this value of x, so this is the

final answer.

Example 5.3. Solve the equation 3 5

4 5 5

x

x x+ =

− −.

Solution. For the denominators 4 and x – 5, the LCD = 4(x – 5).

Complements for denominators to the LCD are:

For 4, the complement is x – 5.

For x – 5, the complement is 4.

The equation becomes

3( 5) 5 4 4x x− + = .

We solve it:

3 15 20 4 3 5 4 3 4 5x x x x x x− + = + = − = − ,

5 5x x− = − = .

For x = 5, the denominator x – 5 becomes zero, so we reject the value x = 5. Because this

is the only possible solution, the original equation has no solution at all.

Example 5.4. Solve the equation 2

3 4 3

2 16 2 8

x

x x+ =

− −.

Solution.

1) To find the LCD, factor the second and third denominators.

Second denominator:

2 16 ( 4)( 4)x x x− = − + .

Third denominator:

43


2 8 2( 4)x x− = − .

The LCD of all three denominators is 2( 4)( 4)x x− + .

2) Find the complement for each denominator to LCD:

For 2, the complement is 2( 4)( 4) 16x x x− + = − .

For 2 16 ( 4)( 4)x x x− = − + , the complement is 2.

For 2 8 2( 4)x x− = − , the complement is 4x + .

3) Multiply each numerator of the original equation by the corresponding complement,

and drop the denominator. The equation becomes

23( 16) 4 2 3 ( 4)x x x− + = + .

4) Solve the above equation:

2 2 2 23 3 16 4 2 3 3 4 3 48 8 3 12x x x x x x− + = + − + = + ,

2 2 40 103 3 12 48 8 12 40

12 3x x x x x− − = − − = = − = − .

5) Looking at the denominators of the original equation, you may see that the second

denominator 2 16 ( 4)( 4)x x x− = − + . It becomes zero if 4x = . The third

denominator 2 8 2( 4)x x− = − . It becomes zero if 4x = . Therefore, none of the

denominators is zero if 10

3x = − and this is the solution.


1 18

3 2 6

x x

x x x x

−+ =

− + − −.

Solution.

1) To find the LCD, factor the third denominator: 2 6 ( 3)( 2)x x x x− − = − + . You may

notice that the first factor x – 3 is the denominator of the first fraction, and the second

factor x + 2 is the denominator of the second fraction, so 2LCD 6 ( 3)( 2)x x x x= − − = − + .

2) Find the complement for each denominator to the LCD:

For x – 3, the complement is x + 2.

For x + 2, the complement is x – 3.

For 2 6x x− − the complement is 1.



( 2) 1 ( 3) 18x x x x+ + − = − .


2 22 3 18 3 3 18x x x x x x x+ + − = − + − = − ,

44


2 23 3 18 0 4 21 0x x x x x+ − − + = + − = .

The last equation is a quadratic equation. We can solve it by factoring:

( 7)( 3) 0 7x x x+ − = = − and 3x = .

5) Check whether these numbers make any of the denominators of the original equation

zero. The number –7 does not, but 3 does, so we must reject the number 3.

Final answer: the original equation has only one solution x = –7.


3 2 6

4 20 3 2 15n n n n− =

+ − + −.

Solution.

1) To find LCD, first we factor denominators that are factorable. It is possible to factor

the first and third denominators:

4 20 4( 5)n n+ = + and 2 2 15 ( 5)( 3)n n n n+ − = + − .

2) Now we construct the LCD by multiplying all of the above factors (taking the common

factor n + 5 only one time):

LCD 4( 5)( 3)n n= + − .

3) Find the complement for each denominator to the LCD:

For 4 20 4( 5)n n+ = + , the complement is n – 3.

For n – 3, the complement is 4( 5) 4 20n n+ = + .

For 2 2 15 ( 5)( 3)n n n n+ − = + − , the complement is 4.



3( 3) 2(4 20) 6 4n n− − + = .


3 9 8 40 24 5 49 24 5 24 49n n n n− − − = − − = − = + ,

735 73

5n n− = = − .

6) None of the denominators of the original equation is zero for this value of n; so, it is

the solution (check this yourself).

In conclusion consider an equation that has the form of equality of two fractions. Such an

equation is called a proportion. Of course, it can be solved in the same way as before, using

LCD. In some cases, it is more convenient to use an important property of proportion: the

cross-multiplication rule. This rule means the following:

if a c

b d= then ad bc= .

45


In words: the product along one diagonal (a times d) is equal to the product along another

diagonal (b times c).

Example 5.7. Solve the equation 5 2

2 4 3 5x x=

− +.

Solution. The equation is written in the form of a proportion, and we can use the cross-

multiplication rule. We have

5(3 5) 2(2 4)x x+ = − ,

15 25 4 8 15 4 8 25 11 33 3x x x x x x+ = − − = − − = − = − .

None of the denominators of the original equation is zero for this value of x, so the final

answer is x = – 3.

46


Exercises 5

In exercises 5.1 – 5.14, solve the given equations.

5.1. 2 5 3

8 8 6

x x− += −

5.2.

2 7 5

6 6 9

x x+ −= −

5.3. 2 3 5

4 4 6x x+ =

5.4.

7 5 3

8 12 8x x+ =

5.5. 8 3

34 4x x− =

− −

5.6.

9 54

6 6x x− =

− −

5.7. 7 5

24 3x+ =

−

5.8.

7 64

2 5x+ =

−

5.9. 3 7 6

4x x x− =

−

5.10.

7 3 2

5x x x− =

−

5.11. 2

2 42

9 9

x

x x x− =

− −

5.12.

2

3 23

4 4

x

x x x− =

− −

5.13. 2

4 7 4

5 9 5 15

x

x x+ =

− −

5.14.

2

5 6 5

7 4 7 14

x

x x+ =

− +

5.15.

2

1 40

4 5 20

x x

x x x x

−+ =

− + + −

5.16.

2

1 30

6 2 8 12

x x

x x x x

−+ =

− − − +

5.17.

2

9 4 7

5 15 2 6m m m m− =

+ − + −

5.18.

2

8 5 7

3 12 3 12m m m m− =

+ − + −

5.19. 2 3

7 9 4 6x x=

− +

5.20.

4 7

5 3 8 3x x=

+ −

Challenge Problems

5.21. Consider the equation 1

( )( )

x a b

x a x b x a x b

−+ =

+ + + +

47


Prove the following statements

a) If 1a = or 1b = , then the equation does not have solutions.

b) If 1a and 1b , then the equation has the only solution 1x = − .

5.22. Consider the equation

1 ( )

( )( )

x a a b

x a x b x a x b

−+ =

+ + + +


a) If 1a = or 2 1b a= − , then the equation does not have solutions.

b) If 1a and 2 1b a − , then the equation has the only solution 1x a b= − − .


1 2

( )( )

x a b x

x a x b x a x b

− −+ =

+ + + +


a) If 2a = or 2b = , then the equation does not have solutions.

b) If 2a and 2b , then the equation has the only solution 2x = − .


1 ( 1)

( )( )

x a a b x

x a x b x a x b

− − −+ =

+ + + +


a) If 2a = or 2 2b a= − , then the equation does not have solutions.

b) If 2a and 2 2b a − , then the equation has the only solution

2x a b= − − .

In problems 5.25 – 5.28, you may use the following

Hint: m grams of a% solution contains 100

ma grams of pure substance.

5.25. A solution of antifreeze contains 20% alcohol. How much pure alcohol must be

added to 6 gallons of the solution to make a 40% solution? 5.26. How many gallons of a 15% sugar solution must be mixed with 6 gallons of a 40%

sugar solution to make a 30% sugar solution? 5.27. A chemist mixed 4 liters of 18% acid solution with 8 liters of 45% acid solution

What percent of acid is in the mixture? 5.28. Nick mixed 9 oz of apple drink with 8 oz of 48% carrot drink. Find the percent of

pure apple juice in the apple drink if the mixture contained 30% fruit juice.

48

Session 6: Radicals and Fractional Exponents

Session 6

Radicals and Fractional Exponents

Definition of Radicals

Suppose we want to construct a box in the shape of a cube having the volume of 8 3cm .

The problem is to find its dimensions (i.e. the length of its edges). If we denote this length

by x, then the volume is 3x . So, to find x, we need to solve the equation 83 =x . Perhaps

you already recognized that 32 8= , so x = 2 cm is the solution. But what if the volume of

a cube is 6? In this case we need to solve the equation 3 6x = . Similar equations may appear

in different problems, and we need a way to refer to numbers that their cubes (or other

powers) are equal to given numbers. It would be a good idea to invent a special notation

for their solutions. Let’s consider more general equation axn = , in which a and n are

given, and we need to find x. The following symbol was invented to describe the solution

of this equation: n . This symbol is called a radical or n-th root or root of the degree n.

Using it, the solution of the equation axn = can be written as n ax = , so ( )n

n a a= . We

may say that to find x, we take the thn root of the number a. The number n is called the

degree or order of the root. For example, we can read 3 8 as “root of the 3rd order of 8”,

or “3rd root of 8”, or “cube root of 8” (based on the above example with the volume of a

cube). By definition, 3 8 2= . For the solution of the equation 3 6x = , we write 3 6x = .

Note. One might think that 3 6 does not exist. However, this is true only if we think about

integers. Actually, the number 3 6 exists, but it is neither an integer nor a fraction (rational

number). The existence of this number is clearly seen if we want to find the length of a side

for a cube with the volume is 6 (such cube exists, so its sides have some length). Numbers

like 3 6 are called irrational numbers and they cannot be written as fractions. If we want

to find such a number as a decimal, we can only get its approximation accurate to a certain

number of digits after decimal point. For example, using a calculator, we can find that 3 6 1.8 or 3 6 1.82 and so on. Simple expression 3 6 replaces all possible

approximations with any number of digits, and we say that this is “exact value”.

The case when the degree is n = 2 is of special interest as it appears most often. For

example, assume that we want to construct a square with the area of 9, and we are interested

in the length of its side. If we denote this length by x, then the area is 2x . To find x, we

need to solve the equation 92 =x . And the solution is 392 ==x . It is the convention to

drop number 2 in the radical 2 , and we simply write . So, 39 = . We read the

expression 9 as “radical 9” or “square root of 9” (based on the example with the area of

a square).

Note. Formally (algebraically) speaking, the equation 92 =x has two solutions: 3=x and

49


3−=x since the square of both numbers is 9. However, by definition, the radical

always refers to a nonnegative number, so 39 = , not –3. Both solutions of the equation

92 =x can be written as 39 ==x and .39 −=−=x

Similar, if a is a nonnegative number, by n a we always mean a nonnegative number.

Note. If we have the expression 2x and is it not given that x is a nonnegative number,

we may not always write that 2x x= since this equality is wrong when x is negative. In

this case 2x x= − (note that x− is positive if x is negative). For example,

( ) ( )2

3 3 3− = − − = . The correct equality, which is true for any x, is this:

2x x= ,

where x is the absolute value of x. In the above example, ( )2

3 3 3− = − = . In most parts

of this textbook, we will assume that variables (letters) are nonnegative numbers, so we

will omit absolute values.

Now, let’s give the formal definition of the thn root.

Definition. Let a be a nonnegative number, and n be a positive integer. Then the thn root

of a, denoted by n a , is the nonnegative solution of the equation axn = .

In other words, n a is the nonnegative number whose nth power is a:

( )n

n a a= .

Number a, which is inside the radical n , is called the radicand.

Note. We defined n a only for nonnegative a. But what if a is a negative? In this case, we

can also define n a , but only if the degree n is an odd number. For example, 283 −=− ,

since 8)2( 3 −=− . If n is even and a is negative, then thn root n a does not exist as a real

number. For example, 9− does not exist because there is no (real) number x such that

92 −=x .

If the square root of a number is an integer, we call such number a perfect square. For

example, 9 is a perfect square, but 8 is not. To get a list of all perfect squares, we can take

a list of integers 0, 1, 2, 3, …., and square these numbers. We will get the list of perfect

squares: 0, 1, 4, 9, …

Properties of Radicals

Similar to exponential expressions, we can multiply and divide radical expressions with

the same degree very easily. We will assume in this session that a and b are any nonnegative

numbers, and n is any positive integer.

50


Product Rule: nnn abba = .

So, to multiply radical expressions of the same degree, just combine them in one.

Proof. It is enough to show that the thn power of the left and right sides in the Product

Rule is the same. By the definition of the thn root, ( )n

n a a= , ( )n

n b b= , and

( )n

n ab ab= . Using Product Rule for exponents (see Session 1), we have

( ) ( ) ( ) ( )n n n n

n n n n na b a b ab ab = = = .

Example 6.1. Multiply and simplify 2 18 .

Solution. 2 18 2 18 36 6. = = =

Quotient Rule: nn

n

b

a

b

a= .

As you see, similar to the Product Rule, we can combine two radical expressions of the

same degree into one. The proof is also similar.

Note. Informally, you may think about radicals as umbrellas above the numbers. Product

and Quotient Rules say that you may replace two umbrellas with one that covers both

numbers.

Example 6.2. Divide and simplify 50

2.

Solution. 50 50

25 5.22

= = =

Below we will show how to combine radical expressions of different degrees.

Power Rule: ( )m

n mn a a= for positive integer m.

In particular, ( )n

n nn a a a= = (this is actually a definition of the radical).

Proof. By the definition of the thm power and the above Product Rule,

( )copies

... ...m

n mn n n n n

m

a a a a a a a a= = =

Example 6.3. Simplify ( )4

2 .

Solution. ( ) ( )4 4

2 2 16 4.= = =

51


Fractional Exponents

So far, we multiply and divide radical expressions of the same degree only. But what if we

need to operate with different degrees? For example, is it somehow possible to represent

the product 3 22 using only one radical? Here we learn how to do this. The idea is to

set up the connection between radicals and exponents. In this way, we could apply product,

quotient and power rules of exponents to radicals.

Let’s try to represent radical expression n a as exponential one with the base a and some

power m: mn aa = . If we raise both sides of the equation to the thn power, we will have

( ) ( ) nmnmn

n aaa == . But ( ) 1aaan

n == . Therefore, nmaa =1 .

From here we can equate powers: 1 = mn, and n

m1

= . Now the expression mn aa = can

be written as nn aa

1

= . We were able to present the thn root as an exponential expression

with the fractional power n

1! We use this representation as the definition.

Definition. Let a be any nonnegative number, and n be any positive integer. Then

nn aa =

1

.

In particular, 1

2a a= .

We can easily generalize this definition to exponential expressions with arbitrary fractional

power n

m. To do this, just raise both sides of the equation nn aa =

1

to the thm power and

use the power rules for exponents and radicals:

( )1

mm

mn mnn na a a a

= =

.

This motivates the following

Definition. Let a be any non-negative number, and n

m be any positive fraction. Then

n mn

m

aa = .

Note. Be careful to put m and n on the right side in the correct places: the numerator m is

the power of a, and n is the degree of the radical.

Representing radicals using fractional exponents extends our ability to manipulate with

radicals. Here is an example of how it may help to simplify radical expressions of different

degrees.

Example 6.4. Simplify the expression 32 2 (combine using one radical).

Solution.

1 1 1 3 2 516 53 63 2 3 6 622 2 2 2 2 2 2 2 32

++

= = = = = = .

52


In session 3 we considered exponential expressions with negative integer exponents. In this

session we dealt with positive fractional exponents. Now we can combine both together

using the corresponding formulas

1k

ka

a

− = and n mn

m

aa = .

From here we see the connection between negative fractional exponents and radicals:

1 1m

nm n mn

aa

a

−

= = , a > 0.

Note. In the definition of the expression 𝑎𝑚

𝑛 with fractional exponent n

m, where m and n

are integers, we always assume that the base a is a positive number. The reason is to

avoid ambiguous situation that may arise when a is negative. For example, compare the

following expressions

(−8)13 and (−8)

26.

Since 1

3=

2

6, it is reasonable to expect that both expressions are equal. However,

(−8)13 = √−8

3= −2, but (−8)

26 = √(−8)26 = √64

6= 2.

Example 6.5. Simplify the expression

10 7

5

3

3.

Solution.

77 1 7 2 5 110 7 10

10 5 10 10 215

5

3 33 3 3 3 3

33

−−

= = = = = = .

Simplification of Square Roots

Here we focus on square roots, but a similar technique can also be used for general

exponents. By simplification we mean a modification of given expression in such a way

that inside the radical remains a smallest possible expression (simplest radicand).

1. Simplification of the numerical expression a (a is a number).

Let’s re-write the Product Rule, described above, from right to left: nnn baab = . In this

way we may split one radical expression into a product of two. In particular,

√𝑎𝑏 = √𝑎 × √𝑏. This property helps to simplify some radical expressions.

Example 6.6. Simplify 12 .

Solution. We split 12 as 4 3 . Then 12 4 3 4 3 2 3= = = .

53


Note. We can also split 12 in a different way: 12 2 6= . However, this way will not lead

to simplification since both factors, 2 and 6, are not perfect squares. Try to split the given

expression in such a way that one of the factors is a perfect square.

Example 6.7. Simplify 48 .

Solution. It is possible to split number 48 as a product of two factors, one of which is a perfect square, in two ways: 12448 = and 31648 = . Both options work. However, if

we use the first way, we need to continue factoring 12 as 34 , and this way is longer. The

second way is shorter. In general, try to split the given number in such a way that a factor

which is not a perfect square cannot be factored further to contain another perfect square.

So, in this example we choose the second factorization 31648 = since number 3 has no

perfect square factor. Using this, we get 3431648 == .

In sections 2 and 3 below, we assume that 𝑥 ≥ 0.

2. Simplification of nx for even n.

We can represent the square root using an exponent of 1

2, and then use the power rule:

( )1

22

n

n nx x x= = .

We come up to the following method, how to simplify the square root from nx with even

power n: divide power by 2 and remove radical.

Example 6.8. Simplify √𝑥10.

Solution. According to the above rule,

10

10 52x x x= = .

Example 6.9. Simplify 16x .

Solution. According to the above rule, 16

16 82x x x= = .

Note. Do not be misled that in the above example power 16 is a perfect square, so one

might think of taking the square root of 16. Do not take the square root of the exponent,

instead, divide the exponent by two.

3. Simplification of nx for odd n.

Any odd number can be written as 2m + 1, where m is an integer, so we can write nx as

2 1mx + . To simplify this expression, we represent 2 1mx +

as 2mx x (separate even power

and single x). Then 2 1 2 2m m m mx x x x x x x+ = = = .

54


We come up to the following rule how to simplify the square root of 2 1mx + : detach

(separate) the variable x from 2 1mx + : 2 1 2m mx x x+ = , remain x inside the radical, and take

the square root from 2mx which is mx .


Solution. Using the above rule, 7 6 6 3x x x x x x x= = = .


Solution. Using the above rule, 25 24 24 12x x x x x x x= = = .

Similar to the above note for the even power, do not take square root from 25.


Solution. Using the above rule, 9 8 49 9 3x x x x x= = .

Note that number 9 here is used in two different ways depending on whether it is a

coefficient next to a variable or an exponent: take square root from the coefficient but not

from the power.

In conclusion, consider an example that combines all three types of radicals described

above: numbers, and exponential expressions with variables that have even and odd

powers.

Example 6.13. Simplify the expression 12 74 150x y .

Solution.

1st method. (Simplify each factor separately). The radicand is a product of three factors:

number 150, 12x and

7y . We can simplify each of them separately:

65625150 == ,

612 xx = ,

yyyyy 367 == .

From here, 12 7 12 7 6 3 6 34 150 4 150 4 5 6 20 6x y x y x y y x y y = = = .

2nd method. (Simplify the entire expression at once).

12 7 12 6 6 3 6 34 150 4 25 6 4 5 6 20 6x y x y y x y y x y y = = = .

55


Exercises 6

In exercises 6.1 and 6.2, evaluate the given expressions without using a calculator.

6.1. a) 2 8

b) 27

3

c) ( )4

5

d) 1

216

e) 1

38−

f) 3

225

g) 2

327−

h) 3 64 4

6.2. a) 3 27

b) 80

5

c) ( )4

6

d) 1

327

e) 1

236−

f) 2

38

g) 3

24−

h) 6 1281 81

In exercises 6.3 and 6.4, write the given expressions using one radical symbol and simplify.

6.3. 7

914

a

a

6.4.

6 5

3

b

b


6.5. 50 6.6. 54

6.7. 32 6.8. 72

6.9. a) 8x

b) 7x

6.10. a) 6x

b) 5x

6.11. a)

3636y

b) 99y

6.12. a) 4949z

b) 6464z

6.13.

10 5 163 75x y z 6.14. 14 9 135 75x y z

In exercises 6.15 and 6.16, simplify and write your answers using positive exponents only.

6.15.

123/ 4

4/3

a

b

−

−

6.16.

155/3

3/5

b

a

−

−

56

Session 7: Multiplication, Addition and Subtraction of Radical Expressions

Session 7

Multiplication, Addition and Subtraction of Radical

Expressions Multiplication of Radical Expressions

In previous session we considered the product rule for radical expressions of the same

degree n: nnn abba = . This rule allows to combine a product of two (or more) radical

expressions into one. Here we consider more examples for multiplication of radical

expressions. We restrict ourselves to square roots only. Also, we assume that all letters

represent nonnegative numbers.

Let’s recall that a a a = . This simple formula allows us to avoid tedious calculations in

some cases: if you notice inside the radical repeated factors, do not multiply them, just

extract the common factor from the radical.

Example 7.1. Multiply and simplify 23 46 .

Solution. One way to proceed is to directly multiply the radicands:

23 46 23 46 1058 = = .

Now you need to simplify 1058 . Even though it is possible, this is not the best way since

it might be difficult to continue. Notice, however, that 46 23 2= , and it is much easier to

proceed like this:

23 46 23 23 2 23 23 2 23 2 = = = .

Example 7.2. Multiply and simplify ( )( )5 3 45 7 3 14x y x y

Solution. We multiply separately numbers which are outside the radicals (5 and 3) and

radicands (we also represent 14 as 7 2 ):

( )( ) ( ) ( ) ( )5 3 4 5 3 4

8 4 4 2 4 2

5 7 3 14 5 3 7 7 2

15 7 7 2 15 7 2 105 2

x y x y x y x y

x yy x y y x y y

=

= = =

Note that the radicand 2y doesn’t have perfect square factors, so we can not simplify the

expression further.

Addition and Subtraction Radicals

Contrary to the multiplication rule a b ab = , there is no simple rule to add or subtract

57


radical: in general, a b a b+ + . Here is an example:

9 16 3 4 7+ = + = , but 9 16 25 5+ = = , so 9 16 9 16+ + .

We can add or subtract radicals directly only if the radicands are exactly the same. This

procedure is similar to combining like terms.

Example 7.3. Add 5 7 3 7+ .

Solution. Similar to combining like terms: 5 3 8x x x+ = , we have 5 7 3 7 8 7+ = .

Example 7.4. Simplify the expression 5 5a x b x+ .

Solution. 5 5 ( ) 5a x b x a b x+ = + .

Example 7.5. Simplify the expression 2 3 5 4 2 5x y z y x z+ − − .

Solution. This expression contains four terms. As we indicated above, we can combine

only those terms that have the same radicands. We can combine the first term with the third

(they contain the same 2y ), and the second with the fourth (they contain the same

5z ):

( ) ( )2 3 5 4 2 5 4 2 3 5x y z y x z x y x z+ − − = − + − .

Note. As you can see, the final answer contains two radicals. We cannot combine them in

one radical because they have different radicands, so no further processing possible.

There are cases when, even if the original expression contains different radicands, it is still

possible to combine them. These are the cases when after simplifying the individual

radicals, they have common radicands.

Example 7.6. Simplify the expression

a) 320 45u u u+ .

b) 6 8 5 27 4 32 2 75+ − −

Solution.

a) There are two radicands here and they are different, so we cannot combine them initially.

Let’s simplify them first (we will process the entire expression):

3 220 45 4 5 9 5 2 5 3 5 5 5u u u u u u u u u u u u u+ = + = + = .

b) There are four radicals here and all of them have different radicands. Similar to part a),

we simplify them first:

58


6 8 5 27 4 32 2 75 6 4 2 5 9 3 4 16 2 2 25 3

6 2 2 5 3 3 4 4 2 2 5 3 12 2 15 3 16 2 10 3.

+ − − = + − −

= + − − = + − −

Now, the first and the third terms contain the same 2 , and the second and the fourth

contain the same 3 . Therefore, we can combine them:

( ) ( )12 2 15 3 16 2 10 3 12 16 2 15 10 3 4 2 5 3+ − − = − + − = − + .

It is not possible to combine further since both radicands are different and cannot be

simplified more.

Mixed Problems

Example 7.7. Multiply and simplify ( )3 5 2 15 4 30xy x y− .

Solution. We can distribute using the usual distributive property:

( )3 5 2 15 4 30 3 5 2 15 3 5 4 30xy x y xy x xy y− = − .

Next, we can simplify each term separately:

23 5 2 15 3 2 5 15 6 5 15 6 5 5 3 6 5 3 30 3xy x xy x xy x x y x y x y = = = = = ,

23 5 4 30 3 4 5 30 12 5 30 12 5 5 6 12 5 6 60 6xy y xy y xy y xy y x y x = = = = =

.

Finally, we subtract the last expression from the previous and get the answer

( )3 5 2 15 4 30 30 3 60 6 .xy x y x y y x− = −

Example 7.8. Multiply and simplify ( )( )635423 +− .

Solution. As in Example 7.7, we start with distributing:

( )( ) 643546233523635423 −−+=+− .

Again, we can simplify each term separately (if you want, you may simplify them

simultaneously):

3 2 5 3 3 5 2 3 15 6 = = ,

3 2 6 3 2 6 3 2 2 3 3 2 3 6 3 = = = = .

320354 = .

From here,

59


( )( ) 6432036615635423 −−+=+−

= (15 4) 6 (6 20) 3 11 6 14 3− + − = − .

Below, we consider an example that can be easily solved using the following difference of

squares formula, which we already mentioned in session 2:

22))(( bababa −=+− .

Example 7.9. Multiply and simplify ( )( )73527352 +− .

Solution. We can solve this problem in the same way as we did in Example 7.8 by

distributing. Notice, however, that inside the first and the second sets of parentheses we

have difference and sum of the same expressions 2 5 and 3 7 respectively. Therefore,

we can use the above difference of squares formula with 52=a and 73=b . We have

( ) ( ) 205452522

22

2 ====a ,

and

( ) ( ) 637973732

22

2 ====b .

Now, subtract 2 2a b− and get the answer

( )( )2 5 3 7 2 5 3 7 20 63 43− + = − = − .

In the following example we solve a problem similar to Example 7.9 in more general

form.

Example 7.10. Multiply and simplify ( )( )yxnmyxnm +− .

Solution.

( )( ) ( ) ( )

( ) ( )

2 2

2 22 2 2 2 .

m n x y m n x y m n x y

m n x y m n x y

− + = −

= − = −

Note. The expressions yxnm − and yxnm + are called conjugate to each other.

Example 7.10 shows that the product of conjugate expressions does not contain radicals.

In the next session, we will use this property to “rationalize” the denominators.

60


Exercises 7

In exercises 7.1 – 7.6, multiply and simplify the given expressions.

7.1. a) 7 7

b) 2018 2018

c) 47 94

7.2. a) 5 5

b) 2019 2019

c) 26 78

7.3. 14 21 7.4. 18 30

7.5. a) 5 6 3 33 15p q p q

b) ( ) ( )3 2 74 24 6 72a b a b

7.6. a) 8 3 3 55 15m n m n

b) ( ) ( )9 6 32 21 7 42u v u v

In exercises 7.7 and 7.8, add or subtract the given expressions.

7.7. a) 3 6 8 6+

b) 3 7 5 7−

7.8. a) 7 5 2 5+

b) 5 3 8 3−


7.9. 3 6 2 7 4 7 5 6m n k k m n− + + 7.10. 7 3 2 6 4 6 5 3p q r r p q− + +

7.11. a) 8 50+

b) 5 27 2 48−

c) 8 24 4 20 2 54 6 45+ − −

7.12. a) 12 27+

b) 6 45 4 80−

c) 6 63 7 48 5 28 2 108− + −

In exercises 7.13 – 7.26, multiply and simplify the given expressions.

7.13. ( )10 2 5+ 7.14. ( )6 2 3+

7.15. ( )5 7 4 63 3 7− 7.16. ( )4 3 6 3 2 12−

7.17. ( )( )2 5 3 6 4 5 2 6+ − 7.18 ( )( )5 2 4 7 3 2 2 7− +

7.19. ( )( )3 6 2 3 3 6 2 3− + 7.20. ( )( )4 7 3 5 4 7 3 5+ −

7.21. ( )( )5 5 2 2 5 5 2 2+ − 7.22. ( )( )6 6 3 3 6 6 3 3− +

61


7.23. ( )2

3 7 2 5+ 7.24. ( )2

2 6 4 3+

7.25. ( )2

3 7 2 5− 7.26. ( )2

2 6 4 3−

Challenge Problems

7.27. Prove that 5 24 2 3+ = + .

7.28. Let’s call a square root simple if there are no other square roots inside it.

Represent the expression 3 8 1+ − as a simple square root.

7.29. Prove that for any nonnegative number a, 1 4 1a a a+ + − = .

62

Session 8: Rationalizing the Denominators and Solving Radical Equations

Session 8

Rationalizing the Denominators and Solving Radical

Equations Rationalizing the Denominators

Similar to the product rule √𝑎 × √𝑏 = √𝑎𝑏, we can use the following quotient rule to

divide radicals (we consider here square roots only): a a

bb= . As with the product rule,

this rule allows to replace two “umbrellas” (two radicals) for a and b, with one “umbrella”

that covers both. For example, 10 10 5

6 36= = . In some cases, it is desirable to modify

expressions like this further to get rid of the radical in the denominator. The procedure to

do this is called rationalizing the denominator.

The general idea to rationalize the denominator is to use the main property of a fraction: if

we multiply the numerator and denominator of a fraction by the same nonzero expression,

the value of the fraction remains the same (even if the fraction will look different). We

consider here two types of fractions: one with a single term with a radical in the

denominator, and another with a sum (or difference) of two terms (where at least one of

them contains radical).

Fractions with a single radical term in the denominator.

Such fractions have the following general form expr

m n, where expr means some expression.

To rationalize the denominator, we multiply both the numerator and denominator by n

and simplify the denominator, using the property n n n = . In doing this, we get rid of

the radical in the denominator (so, we rationalized the denominator):

expr expr exprn n

mnm n m n n= = .

Example 8.1. Rationalize the denominator: 5

3.

Solution. We have 5 5

3 3= . To continue (to get rid of 3 in the denominator), we

multiply the numerator and the denominators by 3 :

5 5 3 15

33 3 3

= =

.

63



5 8.

Solution. Following the same method, we multiply the numerator and denominator by 8

(it is not needed to also multiply by 5):

6 6 8 6 8 6 4 2 6 2 2 3 2

5 8 5 8 5 8 105 8 5 8 8

= = = = =

.

Here we also simplified 8 . If you compare initial fraction 6

5 8 with the final answer

3 2

10, you see that they look completely different. However they have exactly the same

numerical values (you can check this using a calculator).

Fractions with the sum or difference of two terms in the denominator.

Such fractions have the following general form

expr

m n x y− or

expr

m n x y+,

where expr, as before, means some expression. The denominators in these two fractions

are conjugate to each other. As we already saw in example 7.10 from the previous session,

their product is a rational expression (i.e. an expression with no radicals):

( )( ) 2 2m n x y m n x y m n x y− + = − .

This property allows us to rationalize the denominators: multiply the numerator and

denominator by the expression conjugate to the denominator.

Note. If you try to multiply the numerator and denominator of the above fractions by only

one of the radicals (for example, by n ), you will still have radicals in denominator. So,

do not confuse the two cases: single term in the denominator (when we multiply the

numerator and denominator by the radical in the denominator) and the sum or difference

of two terms (when we multiply the numerator and denominator by the expression

conjugate to the entire denominator). For example, consider two fraction:

a

b c and

a

b c+.

We use different methods to rationalize them: for the first fraction we multiply the

numerator and denominator by c , while for the second fraction we multiply the

numerator and denominator by b c− :

64


( )( )( )

( )2

,a b c a b ca a c a c a

bc b cb c b c c b c b c b c

− −= = = =

− + + −

.


2 3−.

Solution. Here we have a difference of two terms in the denominator: 2 3− . Therefore,

we multiply the numerator and denominator by the expression conjugate to the

denominator. The conjugate expression is 2 3+ .

( )

( )( ) ( )2

2

1 2 31 2 3 2 32 3

4 32 3 2 3 2 3 2 3

+ + += = = = +

−− − + −

.

Final answer looks pretty nice: 1

2 32 3

= +−

.

Example 8.4. Rationalize the denominator: yx

x

23

3

+

−.

Solution. The expression conjugate to the denominator is 3 2x y− . We multiply the

numerator and denominator by it:

( )( )( )( ) ( ) ( )

2 22 2

3 3 2 3 2 3 3 3 23

3 2 3 2 3 2 3 2

3 2 9 6.

9 4

x x y x x x y x yx

x y x y x y x y

x xy x y

x y

− − − − + −= =

+ + − −

− − +=

−

Solving Radical Equations

We now consider equations that contain radicals. They can be transformed into equations

without radicals in two steps:

Step 1: If needed, isolate the radical expression (i.e. leave it alone on one side of the

equation).

Step 2: Square both sides of the equation.

Note. When you square both sides of an equation, it is possible to get so-called “extraneous

solutions”, that is the answers that are not solutions to the original equation. Here is a

simple example: 1 2x + = . Obviously, this equation has only one solution 1x = . Now, if

we square both sides of this equation, we get ( )2 21 2x + = , or

( )( )2 22 1 4 2 3 0 1 3 0x x x x x x+ + = + − = − + = .

65


The last equation has two solutions: 1x = and 3x = − . However, the value 3x = − is not a

solution to the original equation 1 2x + = . The conclusion from this note is this: always

check your final answers with the original equation.

Example 8.5. Solve the equation 2 1 7x − = .

Solution. Here radical is already isolated and the first step is not needed. We just square

both sides:

( )2

22 1 7 2 1 49 2 49 1 2 50 25x x x x x− = − = = + = = .

It is easy to verify that number 25 is a solution of the original equation:

2 25 1 49 7 − = = . Final answer: x = 25.

Example 8.6. Solve the equation 6 5 7 6x − + = .

Solution. Here the radical is not isolated, so we isolate it by moving number 7 to the right

side:

6 5 6 7x − = − or 156 −=−x .

Now, the radical is isolated and we square both sides:

( ) ( )2 2

6 5 1 6 5 1 6 1 5 6 6 1x x x x x− = − − = = + = = .

So, it looks like x = 1 is a solution. Let’s check it with the original equation:

6 1 5 7 1 7 1 7 8 − + = + = + = .

However, the right side of the original equation is 6, not 8, so 1x = is not a solution and

we reject it. The original equation has no solutions at all.

Note. It can be seen at the very beginning without doing anything that the original equation

doesn’t have solutions. Indeed, the square root on the left side is always nonnegative, and

by adding it to 7, we cannot get 6.

Example 8.7. Solve the equation 24 14 3 2 3 0x x x+ + − − = .

Solution. Here again the radical is not isolated, and we isolate it by moving terms 2x and

3 to the right side: 24 14 3 2 3x x x+ + = + .

Now square both sides:

( ) ( )2

22 2 24 14 3 2 3 4 14 3 4 12 9x x x x x x x+ + = + + + = + + .

Reducing (cancel out) 24x from both sides, we get

66


14 3 12 9 2 6 3.x x x x+ = + = =

Finally, we check x = 3 with the original equation:

24 3 14 3 3 2 3 3 36 42 3 6 3 81 9 9 9 0 + + − − = + + − − = − = − = .

So, x = 3 is a solution.

Example 8.8. Solve the equation 2 11 4x x+ − = .

Solution. Here the radical is not isolated, and we isolate it by moving x to the right side:

2 11 4x x+ = + . Now we square both sides:

( ) ( )2 2 22 11 4 2 11 16 8x x x x x+ = + + = + + .

We obtained a quadratic equation. Let’s write it in the standard form 2 0ax bx c+ + = . For

this, we first switch left and right sides: 216 8 2 11x x x+ + = + , and then move 2 11x + to

the left:

216 8 2 11 0x x x+ + − − = or

2 6 5 0x x+ + = .

The last equation can be solved by factoring: ( )( )1 5 0x x+ + = . We’ve got two solutions

of the quadratic equation: 1x = − and 5x = − . We need to check them with the original

equation.

1x = − : 2 11 2 ( 1) 11 ( 1) 9 1 3 1 4x x+ − = − + − − = + = + = . So, 1x = − is a solution.

5x = − : 2 11 2 ( 5) 11 ( 5) 1 5 1 5 6 4x x+ − = − + − − = + = + = . So, 5x = − is not a

solution and we reject it.

Final answer: the original equation has only one solution 1x = − .

Example 8.9. Solve the equation 2 6 3x x− + = .

Solution. Here the radical is not isolated, and we isolate it by moving 3 to the right side:

2 6 3x x− = − . Now we square both sides:

( ) ( )2 2 2 2 22 6 3 2 6 6 9 0 6 9 2 6 0 8 15x x x x x x x x x x− = − − = − + = − + − + = − +

( )( )2 8 15 0 3 5 0x x x x − + = − − = .

We’ve got two solutions of the quadratic equation: 3x = and 5x = . We need to check

them with the original equation.

3: 2 6 3 2 3 6 3 0 3 3x x= − + = − + = + = . So, 3x = is a solution.

5 : 2 6 3 2 5 6 3 4 3 2 3 5x x= − + = − + = + = + = . So, 5x = is a solution.

Final answer: the original equation has two solutions, 3x = and 5x = .

67


Exercises 8

In exercises 8.1 – 8.16, rationalize the denominator and simplify.

8.1. 5

6

8.2.

4

3

8.3. 3

5

8.4.

6

7

8.5. 2

3 20

8.6.

3

5 27

8.7. 4

5 3

8.8.

2

3 5

8.9. 4

5 3+

8.10.

2

3 5+

8.11. 1

2 3−

8.12.

1

5 6+

8.13. 1

6 3+

8.14.

1

6 5−

8.15. 2

4 3

u

u v

+

−

8.16.

2 5

a b

a b

+

+

In exercises 8.17 – 8.28, solve the given equation.

8.17. 3 4 5x + = 8.18. 4 3 7x − =

8.19. 5 4 2 8x − + = 8.20. 7 1 4 2x + − =

8.21. 6 8 5 3x − + = 8.22. 8 9 7 2x + + =

8.23. 216 70 7 4 7 0x x x+ − − − = 8.24. 225 12 7 5 1 0x x x− + − + =

8.25. 29 7 18 4 3x x x+ − − = 8.26. 236 19 25 3 6x x x− − + =

68


8.27. a) 3 10 2x x+ − =

b) 3 7 3x x+ − =

8.28. a) 4 13 4x x+ − =

b) 4 21 4x x+ − =

Challenge Problems

8.29. Consider the equation 2ax b x b+ − = with a > 0 and b > 0.

Prove the following statements:

1) If a < b, then the equation has the only solution x = 0.

2) If a b , then the equation has two solutions x = 0 and x = a – 2b.

8.30. Consider the equation 2ax b ar x b r+ − − = − with a > 0.


1) If b < 0, then the equation has only one solution x = a + r – 2b.

2) If b > a, then the equation has only one solution x = r.

3) If 0 b a , then the equation has both the above solutions.

Hints: 1) Check that the quadratic equation ( )22ax b ar x b r+ − = + − has the solutions

x = r and x = a + r – 2b.

2) Use the property 2

, if 0,

, if 0.

u uu u

u u

= =

−

69

Session 9: Complex Numbers and Squared Form of Quadratic Equations

a a i− =

Session 9

Complex Numbers and

Squared Form of Quadratic Equations

In section 2 we considered quadratic equations that can be solved by factoring. However,

not every quadratic equation can be solved by factoring (using rational numbers). Below

we consider corresponding examples. Also, unlike linear equations, quadratic equations do

not always have real solutions. We will discussion this case first.

Complex Numbers

A simple example of an equation without real roots is the equation 012 =+x . It can be

written as 12 −=x . Obviously, this equation does not have solutions because the square of

a real number cannot be negative.

However, it is possible to introduce a special symbol that can be treated as a solution of the

equation 012 =+x . Usually, this symbol is denoted by the letter i and is called the

imaginary unit (that’s why the letter i). Of course, i is not a real number. It has the property

that 12 −=i . Also, we can write that 1−=i .

Note. You may be disappointed with such a “definition” of the number i. Indeed, it looks

like we introduce an object that does not exist: the equation 012 =+x does not have any

real solutions, and we use letter i for non-existent solution. If you have such feelings, you

are not the only one. For more than two hundred years many mathematicians felt the same

way. Only in the 18th century the exact theory of so-called complex numbers was created

which included the symbol i as well as other related to it “magic” numbers. We can look

at a complex number as an ordered pair of two real numbers.

Definition. A complex number z is an expression of the form z = a + bi. Here a and b are

two real numbers, and i is a symbol with the property 12 −== iii . Actually, we are saying

that ii is equal to –1 by definition of i. The symbol i is called the imaginary unit, the

number a is the real part, and the number b is the imaginary part of the complex number

z. The form a + bi is called the standard form of a complex number.

Using the symbol i, a square root of any negative number can be written as a complex

number: if a is a non-negative real number, then 1a a a i− = − = , so

, 0a .

Example 9.1. Express 48− in terms of i and simplify.

Solution. 48 48 16 3 4 3i i i− = = = .

We can operate with complex numbers in the same way as with polynomials or rational

expressions: we can perform all arithmetic operations with them, distribute, combine like

70


terms. The only specific property we use is that if we multiply i by i, the result is –1. We

say that two complex numbers ibaz 111 += and ibaz 222 += are equal, if separately their

real and imaginary part are equal: 21 aa = and 21 bb = .

If a is a real number, then it can be written as a complex one via: a = a + 0i. In this way we

can consider the set of real numbers as a subset of the set of complex numbers. In other

words, the set of complex numbers can be treated as an extension of the set of real numbers.

Let’s consider arithmetic operations with complex numbers.

Example 9.2.

1) Add (3 + 2i) + (1 + 5i).

2) Subtract (6 + 3i) – (4 – 7i).

3) Multiply (1 + i)(2 + 3i).

4) Multiply (3 + 4i)(3 – 4i).

Solution. As we mentioned above, we can perform these operations as with ordinary

algebraic expressions just keeping in mind that 12 −=i .

1) (3 + 2i) + (1 + 5i) = (3 + 1) + (2 + 5)i = 4 + 7i.

2) (6 + 3i) – (4 – 7i) = (6 – 4) + (3 – (–7))i = 2 + 10i.

3) iiiiiiiiii 513232)1(32323232)32)(1( 2 +−=−++=−+++=+++=++ .

4) We can proceed the same way as in part 3) or use as a shortcut difference of squares

formula that we already used in session 2 (see note after example 2.1): 22))(( bababa −=+− . Using this formula we have

( ) ( ) ( )223 4 (3 4 ) 3 4 9 16 1 9 16 25i i i+ − = − = − − = + = .

Note that the result of multiplication here is a real number. Below we will use a similar

result in a general form.

Next, consider division of complex numbers. At first glance, the quotient of two complex

numbers does not look like a complex number. For example, can the quotient i

i

23

45

+

+ be

represented as a complex number in the standard form a + bi? It turns out, the answer is

yes. The method of dividing complex numbers is similar to the one we used in the previous

session to rationalize the denominator of radical expressions with two terms in the

denominator: we multiplied the numerator and denominator of a fraction by the expression

conjugate to the denominator. Similarly, consider the following pair of complex numbers

1z and 2z :

biaz +=1 , biaz −=2 .

These numbers have the same real part a, while their imaginary parts b and – b differ only

by a sign. Numbers a bi+ and a bi− are called complex conjugate to each other. An

important property is that if we multiply them, the product is a real number:

71


22222

21 ))(( baibabiabiabiabiazz +=−+−=−+= .

This property allows us to represent the quotient bia

dic

+

+ as a complex number in standard

form by multiplying the numerator and denominator of this fraction by a bi− which is the

conjugate of the denominator:

2

2 2 2 2

( )( ) ( ) ( )

( )( )

c di c di a bi ac bci adi bdi ac bd ad bc i

a bi a bi a bi a b a b

+ + − − + − + + −= = =

+ + − + +

So,

iba

bcad

ba

bdac

bia

dic2222 +

−+

+

+=

+

+.

As you see, the result is written as the sum of two parts, the real part being 2 2

ac bd

a b

+

+ and

the imaginary part being 2 2

ad bc

a b

−

+. Therefore, the result of division is a complex number in

standard form. We got the general formula for dividing two complex numbers.

Note. The above formula looks rather complicated. Don’t worry: you do not need to

memorize it. Just keep in mind the method for dividing complex numbers: multiply the

numerator and denominator by the number conjugate to the denominator.

Example 9.3. Divide (5 + 4i) by (3 – 2i).

Solution.

25 4 (5 4 )(3 2 ) 15 10 12 8 (15 8) (10 12) 7 22 7 22

3 2 (3 2 )(3 2 ) 9 4 13 13 13 13

i i i i i i i ii

i i i

+ + + + + + − + + += = = = = +

− − + +.

If the denominator of a fraction contains only the imaginary part (so, the real part is equal

to zero), to divide, simply multiply the numerator and denominator by i.

Example 9.4. Divide 1 by i.

Solution. 1 1

1

i ii

i i i

= = = −

−. This result can also be written as

1i i− = − .

Example 9.5. Divide 7 – 5i by 4i.

Solution. ( ) 2

2

7 57 5 7 5 7 5 5 7.

4 4 4 4 4 4

i ii i i ii

i i i i

− − − += = = = − −

−

Now we consider how to raise the imaginary unit i to any positive integer power.


72


1) 3i 2) 4i 3) 5i 4) 6i

Solution.

1) iiiii −=−== )1(23 . 2) ( ) ( )2 24 2 1 1i i== = − = .

3) ( ) ( )2 25 4 2 1i i i i i i= = = − = . 4) ( ) ( )

3 36 2 1 1.i i= = − = −

The above example suggests a method for calculating ni for any positive integer power n.

First of all, note that ( )1 1k

− = , if k is even, and ( )1 1k

− = − , if k is odd. To calculate ni ,

consider separetely cases of even and odd n.

1) If n is even, it can be written as n = 2k, where k is an integer. Then

( ) ( )2 2 1k kn ki i i= = = − .

From here ni is equal to 1 or –1, depending on whether k is even or odd. For even k,

1ni = , and for odd k, 1ni = − .

2) If n is odd, it can be written as n = 2k + 1, where k is an integer. Then

( ) ( )2 1 2 2 1k kn k ki i i i i i i+= = = = − .

From here ni is equal to i or –i, depending on whether k is even or odd. For even k,

ni i= , and for odd k,

ni i= − .


1) 100i 2) 50i 3)

29i 4) 35i

Solution.

1) ( ) ( )50 50100 2 1 1.i i= = − =

2) ( ) ( )25 2550 2 1 1.i i= = − = −

3) ( ) ( ) ( )14 14 1429 28 2 1 1 .i i i i i i i i= = = − = − =

4) ( ) ( )17 1735 34 2 1 .i i i i i i i= = = − = −

Squared Form of Quadratic Equations

We will call the quadratic equation written in the squared form, if it is written as

rqpx =+ 2)( .

In the next session, we show that any quadratic equation can be transformed to this form.

To solve this equation, we will use the square-root property. It states that the equation 2x c= has two solutions: x c= and x c= − . Using this property, we take the square

root on both sides of the above equation in the squared form. As a result, we get two linear

73


equations: rqpx =+ and rqpx −=+ . It is very common to write both equations as a

single one using the symbol “ ”: rqpx =+ . From here, rqpx −= , and

p

rqx

−= . This is the final answer for the solutions of the equation rqpx =+ 2)( .

Note. Keep in mind that the formula p

rqx

−= represents two solutions:

p

rqx

+−= and

p

rqx

−−= .

Example 9.8. Solve the quadratic equation 5)23( 2 =+x .

Solution. Applying the square-root property, take the square root on both sides using the

symbol “ ”: 523 =+x . From here, 523 −=x and 3

52−=x .

Note. The final answer 3

52−=x represents two exact solutions written in radical

form: 3

52+−=x and

3

52−−=x . If we want to get solutions as decimal numbers, we

can do this only approximately. Using a calculator to approximate 5 as 2.236, we can

get the following approximations for the solutions:

079.03

236.22

3

52

+−

+−=x and 412.1

3

236.22

3

52−=

−−

−−=x .

Example 9.8 shows us that not every quadratic equation can be solved by factoring in terms

of rational numbers.

Example 9.9. Solve the quadratic equation 0)25( 2 =−x .

Solution. By taking the square root on both sides, we get only one linear equation

5 2 0x − = . From here, 5

2=x . So, the given quadratic equation has only one (unique)

solution 5

2=x .

Example 9.10. Solve the quadratic equation 2 5 0x + = .

Solution. Write this equation in squared form 2 5x = − . From here, 5x = − . Number

5− can be written in terms of imaginary unit i: 5 5 i− = , so 5x i= . The solutions

are two imaginary numbers 5 i and 5 i− .

74


Example 9.11. Solve the quadratic equation 07)34( 2 =+−x .

Solution. Write the equation in squared form: 7)34( 2 −=−x . Next, apply the square-root

property (take square root on both sides) to get 4 3 7x − = − . Number 7− can be

written in terms of imaginary unit i: 7 7 i− = and we get the equation 4 3 7x i− =

. From here, 4 3 7x i= , and 3 7

4

ix

= . So, the final answer is represented by two

complex conjugate numbers:

1

3 7 3 7

4 4 4

ix i

+ = = + and

2

3 7 3 7

4 4 4

ix i

− = = − .

75


Exercises 9

In exercises 9.1 and 9.2, represent the given expression in terms of i and simplify.

9.1. a) 25−

b) 32−

9.2. a) 16−

b) 27−

In exercises 9.3 and 9.4, add or subtract.

9.3. a) (7 3 ) (2 5 )i i+ + −

b) (4 6 ) (8 2 )i i− − +

9.4. a) (9 4 ) (3 )i i− + −

b) (6 3 ) (2 8 )i i− − −

In exercises 9.5 and 9.6, multiply the complex numbers.

9.5. a) 3(2 4 )i−

b) 4 (5 2 )i i+

c) (4 5 )(2 3 )i i+ +

d) (6 3 )(2 8 )i i− −

e) (5 3 )(6 5 )i i+ −

f) (3 2 )(3 2 )i i− +

9.6. a) 4(3 5 )i+

b) 7 (6 3 )i i−

c) (8 3 )(2 4 )i i+ +

d) (7 3 )(5 6 )i i− −

e) (4 6 )(7 )i i− +

f) (2 4 )(2 4 )i i+ −

In exercises 9.7 and 9.8, divide. Write the answer in standard form for complex numbers.

9.7. a) (8 2 ) 5i+

b) 3 (2 )i

c) (9 6 ) ( 4 )i i− −

d) ( 4 3 ) (5 6 )i i− + −

e) (6 7 ) ( 3 4 )i i+ − +

f) (3 2 ) (3 2 )i i− +

9.8. a) ( 6 7 ) 2i− +

b) 2 ( 3 )i −

c) (4 3 ) (6 )i i+

d) (5 6 ) (4 3 )i i− −

e) (4 5 ) ( 3 2 )i i+ − +

f) (2 7 ) (2 7 )i i− +

In exercises 9.9 and 9.10, calculate

9.9. a) 2018i

b) 2019i

c) 2020i

d) 2021i

9.10. a) 2023i

b) 2024i

c) 2025i

d) 2026i

76



9.11. a) 2(4 3) 6x − =

b) 2 7 0x + =

c) 2(6 4) 3 0x + + =

9.12. a) 2(5 2) 3x + =

b) 2 3 0x + =

c) 2(7 5) 6 0x − + =

77

Session 10: Completing the Square and the Quadratic Formula

Session 10

Completing the Square and the Quadratic Formula

In this session, we show that any quadratic equation can be written in the squared form

described in the previous session. In this way, we also get the quadratic formula, a formula

that allows to solve any quadratic equation by simply substituting its coefficients into the

formula.

Completing the Square

The procedure for converting the quadratic equation from the standard form

02 =++ cbxax into the squared form rqpx =+ 2)( is called the Completing the Square.

This procedure is based on the following two formulas that we already mentioned in

session 3 – the square of the sum and the square of the difference formulas:

2 2 2( ) 2x p x px p+ = + + (Square of the Sum Formula).

2 2 2( ) 2x p x px p− = − + (Square of the Difference Formula).

We start with a special case: consider how to complete the square for the equation

2 2 0x px c+ + = (a = 1, b = 2p).

We can rewrite it like this (bring c to the right):

2 2x px c+ = − .

Compare the left side of this equation with the square of the sum formula (written from

right to left)

2 2 22 ( )x px p x p+ + = + .

We see that the left side of our equation 2 2x px c+ = − is not the square

2( )x p+ because

the term 2p is missing. To make up for this deficit (to complete the square), we add

2p to

both sides of our equation:

2 2x px c+ = − ,

2 2 22x px p c p+ + = − + ,

( )2 2x p c p+ = − + .

We have completed thus the square and presented the equation in the squared form.

Next, consider the case of reduced equation 2 0x bx c+ + = . By moving c to the right, we

get 2 .x bx c+ = − To make the left side

2x bx+ look like 2 2x px+ , we write the coefficient

b as 22

bb = , so

2

bp = . Then, to complete the square, we add

2

2

2

bp

=

to both sides of

equation: 2x bx c+ = − :

78


2 2 2 22

2 2 2 4

b b b bx bx c x c

+ + = − + + = − +

.

In words, to get the term

2

2

b

that we add to both sides of the equation 2x bx c+ = − , we

divide coefficient b for x by 2 and square it.

For the general case of the equation 02 =++ cbxax with arbitrary 0a , we can take just

one additional step: divide both sides of this equation by the leading coefficient a. Then we

get the reduced equation 02 =++a

cx

a

bx .

Example 10.1. Solve the quadratic equation 0562 =++ xx by completing the square.

Solution. Bring the last term 5 to the right: 562 −=+ xx . Next, divide coefficient 6 next to

x by 2 and square:

2

263 9

2

= =

. Add this 9 to both sides of the equation 562 −=+ xx to

get 2 26 3 5 9 4x x+ + = − + = . Complete the square: ( )

23 4x + = .

We can finish solving the equation by applying the square-root property: 243 ==+x

. From here, 23−=x , and we get two solutions: 123 −=+−=x , and 523 −=−−=x .

Final answer: x = – 1 and x = – 5.

Example 10.2. Solve the quadratic equation 0352 =+− xx by completing the square.

Solution. Bring the last term 3 to the right: 352 −=− xx . Divide coefficient –5 next to x

by 2 and square:

2 25 5 25

2 2 4

− = =

. Add 25

4 to both sides of the equation 352 −=− xx

:

2

2 5 255 3

2 4x x

− + = − +

. Complete the square using the square of the difference formula

4

13

2

52

=

−x . From here, applying the square-root property

2

13

4

13

2

5==−x ,

2

135

2

13

2

5 ==x .

The answer is represented in the radical form that combines two solutions

2

135+=x and

2

135−=x

in one formula.

Example 10.3. Solve the quadratic equation 0253 2 =++ xx by completing the square.

Solution. Here the leading coefficient is not 1, it is 3. We can use the following steps to

solve the given equation.

79


1) Divide both sides by the leading coefficient 3: 03

2

3

52 =++ xx .

2) Bring the last term 3

2 to the right:

3

2

3

52 −=+ xx .

3) Divide coefficient 3

5 next to x by 2 and square:

36

25

6

52

=

:

4) Add the number 25

36 to both sides of the equation

3

2

3

52 −=+ xx :

2

2 5 5 2 25.

3 6 3 36x x

+ + = − +

.

5) Complete the square: 36

1

6

52

=

+x .

To get solutions, take square root on both sides: 6

1

36

1

6

5==+x .

Finally, solve for x: 6

1

6

5−=x . We come up with two solutions:

3

2

6

4

6

1

6

5−=−=+−=x

and 16

6

6

1

6

5−=−=−−=x . Final answer:

3

2−=x and 1−=x .

Note. It would be probably easier to solve the equations in examples 10.1 and 10.3 by

factoring. However, we used the method of completing the square to demonstrate its

universal character: any quadratic equation can be solved by this method, contrary to the

method of factoring that does not work in all cases. In Example 11.3 the solutians are

irrational numbers (since they contain 13 ), and cannot be obtained by factoring the

quadratic equation using rational numbers.

Quadratic Formula

In order to obtain the quadratic formula, we repeat steps that we used in Example 10.3 for

the general equation 02 =++ cbxax , 0a written in standard form.

1) Divide both sides by the leading coefficient a toget the reduced equation:

02 =++a

cx

a

bx .

2) Bring the last term c

a to the right side:

2 b cx x

a a+ = − .

3) Divide coefficient b

a next to x by 2 and square:

2 2

22 4

b b

a a

=

80


4) Add 2

24

b

a to both sides of the equation 2 b c

x xa a

+ = − :

2 22

22 4

b b c bx x

a a a a

+ + = − +

.

5) Complete the square on the left side:

2 2

2

2 2

b b bx x x

a a a

+ + = +

.

On the right side, we can combine fractions by getting the common denominator of 24a :

2

2

2

2

2

2

22

2

4

4

4

4

44

4

4 a

acb

a

bac

a

b

a

ac

a

b

a

c −=

+−=+−=+− .

Now we can write 2 2

2

4

2 4

b b acx

a a

− + =

.

This is the squared form of the equation 02 =++ cbxax . To solve it, we apply the square-

root property:

a

acb

a

acb

a

bx

2

4

4

4

2

2

2

2 −=

−=+

To solve for x, bring the term a

b

2 to the right side:

a

acbb

a

acb

a

bx

2

4

2

4

2

22 −−=

−−= .

We get a formula called the quadratic formula

This formula gives the solutions of any quadratic equation written in standard form

02 =++ cbxax , 0a .

The quadratic formula also allows us to get an idea about the “nature” of the solutions for

quadratic equations. Let’s analyze this. First of all, the quadratic formula represents two

solutions 1x and 2x :

2 4

2

b b acx

a

− −=

81


2

1

4

2

b b acx

a

− + −= and

2

2

4

2

b b acx

a

− − −= ,

written together in one formula, using " " operation.

Also, notice that the main part of this formula is the expression inside square root: 2 4b ac−

. This expression is important and it is given a special name.

Definition. The expression 2 4b ac− is called the discriminant of the quadratic equation

02 =++ cbxax , and is denoted by the letter D: 2 4D b ac= − .

Using the discriminant, the quadratic formula can be written in a slightly simpler form

2

b Dx

a

− = or

12

b Dx

a

− += and 2

2

b Dx

a

− −= .

Note. For the reduced quadratic equation 2 0x bx c+ + = (when a = 1), D , when

D > 0, has the following geometric interpretation: it is the distance between roots 1x and

2x on number line: 1 2 .

2 2 2

b D b D D Dx x D

− + − − +− = − = =

The discriminant is a real number, and it may be positive, negative or zero. Let’s see how

the sign of discriminant affects the roots 1x and 2x .

1) Discriminant is positive: 0D . In this case D is a positive number and the

expressions b D− + and b D− − are two different numbers. Therefore, from the

quadratic formula, the quadratic equation has two roots 1x and 2x which are real

numbers and distinct. Roots 1x and 2x may be rational or irrational numbers.

2) Discriminant is zero: 0D = . In this case roots 12

bx

a

−= and 2

2

bx

a

−= , so roots coincide

and the equation has only one real root 2

bx

a= − .

3) Discriminant is negative: 0D . In this case D is an imaginary number and the

equation has two complex roots 1x and 2x which are conjugate to each other.

As you can see, there are only three options regarding the nature of solutions fo any

quadratic equation: it may have one real solution, two (distinct) real solutions, or two

complex conjugate solutions. The sign of the discriminant allows us to distinguish between

these three cases.

We now consider some examples of using the quadratic formula. You can use either form:

with the discriminant or without it. We will use the discriminant form

2

b Dx

a

− = , where

2 4D b ac= − .

Note. When using the quadratic formula, make sure that the quadratic equation is written

82


in the standard form 02 =++ cbxax (the right side must be zero) to identify coefficients

a, b, and c correctly.

Example 10.4. Solve the quadratic equation 23 6 2x x+ = by using the quadratic formula.

Solution. The equation is not in standard form. To get it in standard form, bring 2 from the

right side to the left: 23 6 2 0x x+ − = . Now identify the coefficients and calculate the

discriminant:

3, 6, 2a b c= = = − , so 2 24 6 4 3 ( 2) 36 24 60D b ac= − = − − = + = .

The discriminant is positive, so our equation will have two distinct real solutions

6 60 6 4 15 6 2 15 3 15

2 2 3 6 6 3

b Dx

a

− − − − − = = = = =

.

Both solutions are irrational numbers because they contain 15 .

Example 10.5. Solve the quadratic equation 24 20 25 0x x− + = by using the quadratic

formula.

Solution. The equation is already in standard form. We have

4, 20, 25a b c= = − = , so 2 24 ( 20) 4 4 25 400 400 0D b ac= − = − − = − = .

The discriminant is zero, so our equation will have only one real solution

( 20) 0 20 5

2 2 4 8 2

b Dx

a

− − − = = = =

.

Example 10.6. Solve the quadratic equation 26 5 2 0x x− + = by using the quadratic

formula.

Solution. The equation is already in standard form. We have

6, 5, 2a b c= = − = , so 2 24 ( 5) 4 6 2 25 48 23D b ac= − = − − = − = − .

The discriminant is negative, so our equation will have two complex conjugate solutions

( 5) 23 5 23 5 23

2 2 6 12 12 12

b D ix i

a

− − − − = = = =

.

83


Exercises 10

In exercises 10.1 and 10.2, fill in blanks to complete the squares.

10.1. a) ( )22 8 __ __x x x+ + = +

b) ( )22 3 __ __x x x− + = −

c) ( )22 __ __x x x+ + = +

d) ( )22 5

__ __4

x x x− + = −

10.2. a) ( )22 6 __ __x x x+ + = +

b) ( )22 9 __ __x x x− + = −

c) ( )22 __ __x x x− + = −

d) ( )22 7

__ __3

x x x+ + = +

In exercises 10.3 and 10.4, solve the given quadratic equation by completing the

square.

10.3. a) 2 4 5 0x x− − =

b) 2 5 6 0x x+ + =

c) 2 7 8 0x x− + =

d) 22 3 9 0x x+ − =

10.4. a) 2 2 8 0x x+ − =

b) 2 9 14 0x x− + =

c) 2 3 5 0x x+ − =

d) 23 7 2 0x x+ + =

In exercises 10.5 and 10.6, without solving the equations, determine the number of real

roots.

10.5. a) 27 3 2 0x x− + =

b) 26 9 4 0x x+ − =

c) 23 6 3 0x x− + =

10.6. a) 24 24 36 0x x+ + =

b) 25 6 3 0x x− + =

c) 28 5 2 0x x+ − =

In exercises 10.7 and 10.8, solve the given quadratic equation via the quadratic

formula.

10.7. a) 25 8 1x x− =

b) 216 24 9 0x x+ + =

c) 24 7 5 0x x− + =

d) 26 17 12 0x x− + − =

10.8. a) 27 6 3x x− =

b) 29 42 49 0x x− + =

c) 26 5 3 0x x+ + =

d) 212 28 15 0x x− + − =

Challenge Problem

10.9. Construct a quadratic equation with real coefficients having the root 2 + 3i. How

many such equations can you construct?

84

Session 11: Parabolas

Session 11 Parabolas

In this session we will relate the quadratic equation 02 =++ cbxax to the quadratic function cbxaxy ++= 2 . In other words, we will focus not on solving the above equation, but rather on the relationship between x and y. The graph of the quadratic function (and the function itself) is called a parabola. Recall that for quadratic equation there are three cases of real solutions: it may have one solution, two solutions, or no solutions at all. As a graph, parabola allows to visualize all three cases as well as some other properties of quadratic function.

Let’s start with the simplest (or basic) parabola 2xy = . Notice, first of all, that this function takes the same values for x and –x since ( ) 22 xx =− . Both points ( )2,x x and

( )2,x x− lie on the graph of parabola and they are symmetrical to each other over the y-axis. Therefore, if we draw this parabola only for positive x, then we can reflect this graph over the y-axis to get the entire picture.

Another simple property is that for positive x, the bigger x, the bigger 2y x= . We say that parabola 2xy = increases (for positive x). However, this function is not linear: its graph is not a straight line. Instead, the graph is a curve. To picture this curve, we can calculate several values of parabola for some values of x. The following table represents one of the possible calculations.

x 0 1 2 3 2xy = 0 1 4 9

(x, y) (0, 0) (1, 1) (2, 4) (3, 9) If we plot points (x, y) and connect them with a smooth curve, we will get the picture: Graph of the parabola 2y x= for nonnegative x.

85


To get the entire parabola (to include negative x), we reflect this graph over the y-axis. Here is the final picture. Let’s observe this graph. Notice that for negative x parabola is going down from left to right (we say that parabola decreases), and for positive x parabola is going up (increases). We say that this parabola opens up (or upward). Also, it has the lowest point (0, 0). This point is called the vertex of parabola.

Now consider the second basic parabola 2xy −= . We do not need any special analysis to graph this function. Point ( )2,x x lies on the graph of parabola 2y x= , and point

( )2,x x− lies on the graph of parabola 2xy −= . These points are symmetrical to each

other over the x-axis. Therefore to get the graph of 2xy −= , we can simply reflect the graph of 2xy = over the x-axis (flip the graph upside-down). We say that parabola

2xy −= opens down (downward). Here is its graph:

Now consider the general quadratic function cbxaxy ++= 2 . It turns out that the shape of its graph is similar to one of the above graphs of 2xy = and 2xy −= (depending on whether the leading coefficient a is positive or negative). To understand why, let’s consider three types of transformations (deformations) of graphs of the functions.

2xy =

2xy −=

86


Namely, assume that we know the graph of )(xfy = , and consider how we can construct graphs of functions )(xafy = , ( )y f x k= + , and )( hxfy += .

1) Compare the following graphs of 2y x= , 22y x= and 212

y x= .

As you see, the vertices of all three parabolas remain at the origin and graphs of 22y x=

and 212

y x= resemble the parabola 2xy = . To get the graph of 22y x= , we stretch the

graph of 2xy = in the y-axis, and to get the graph of 212

y x= we compress the graph of

2xy = in the y-axis. More general: if a > 1, the graph of 2axy = is obtained from the graph of 2xy = by stretching in the y-axis if a > 1 (the graph is getting taller), and by compressing in the y-axis if 0 < a < 1 (the graph is getting shorter). The graph of

2y ax= − is obtained from the graph of 2axy = by reflection over the x-axis.

For an arbitrary function )(xfy = , we can get the graph of )(xafy = for positive a in a similar way: we stretch or compress the graphs of )(xfy = along the y-axis depending on whether a > 1 or a < 1. So, the shape of the graph of )(xaf resembles the graph of

)(xf . The graph of ( )y af x= − is obtained from the graph of )(xafy = by reflection over the x-axis. 2) The graph of ( )y f x k= + is obtained from the graph )(xfy = by its shifting along

the y-axis k units. If k > 0, the graph is shifted up, and if k < 0 – down. So, the transformation ( ) ( )f x f x k→ + does not change the shape of the graph of )(xf , it only changes the position of the graph.

3) Finally, consider the graph of )( hxfy += . This graph is obtained from the graph of )(xfy = by its horizontal shifting along the x-axis by h units. It is important not to

confuse the direction of shifting: to the left or to the right. It may seem that for positive h the graph is shifted to the right, and for negative h – to the left. However,

x

y

22y x=

212

y x=

2y x=

12

1

1

2

•

•

•

87


this is wrong. The correct answer is just the opposite: if h > 0, graph is shifted to the left, and if h < 0, to the right. Here is the reason. For positive h, consider two points

0x and 1 0x x h= − . Point 1x lies on the left of 0x . At point 1x , function )( hxf + takes the value 1 0 0( ) ( ) ( )f x h f x h h f x+ = − + = , which is the same as the value of

)(xf at point 0x . Since 1 0x x< , we have shift to the left. Similar reasoning is true when h is negative.

In case of parabola, we consider ( )2y x h= − . Similar to the above arbitrary function

)(xf , we conclude that the shape of the parabola ( )2y x h= − is exactly the same as

the shape of 2xy = , and only the location is different: if h > 0, ( )2y x h= − is located

h units to the right of 2xy = , and if h < 0, h units to the left. The same thing is true

for functions ( )2y x h= − − and 2xy −= . The vertex of both parabolas ( )2y x h= −

and ( )2y x h= − − has the coordinates (h, 0).

If we combine together the above three transformations, we see that if quadratic function is written in the squared form ( )2y a x h k= − + , then its graph resembles

the basic parabolas 2xy = or 2xy −= . In particular, the vertex of the parabola

( )2y a x h k= + + (i.e. its lowest or highest point) has the coordinates (h, k). The

graph of this parabola can be obtained from the graph of 2y ax= by shifting in vertical direction by k units (up or down) and in horizontal direction by h units (left or right) as described above. The graph is symmetric over the vertical line that passes through its vertex (h, k), so the line x = h is the line of symmetry of the parabola. The parabola opens up if a > 0, and opens down if a < 0.

To graph parabola, which is in the squared form ( )2y a x h k= − + , we can use the following steps.

1) Plot the vertex (h, k). 2) Draw dotted vertical line through the vertex. This is the line of symmetry of the

parabola. 3) Identify whether the parabola opens up or down by looking at the sign of the leading

coefficient a. If a > 0, it opens up, if a < 0, it opens down. 4) Draw the parabola. To be more accurate, you may calculate several values of the

parabola and plot corresponding points. In particular, you may put x = 0 to find y-intercept.

Example 11.1. Graph the parabola ( ) 432 2 ++= xy .

Solution. Let’s follow the above steps. We have a = 2, h = –3, k = 4. 1) Plot the vertex (h, k) = (–3, 4):

88


2) Draw dotted vertical line of symmetry through the vertex (–3, 4): 3) Identify how parabola opens (up or down) looking at the leading coefficient a = 2. It

is positive, so parabola opens up. 4) To more accurately draw the parabola, calculate several values (you may choose any

values of x, we choose –1 and –2):

( ) 124312)1( 2 =++−=−y , so the graph contains the point ( – 1, 12). ( ) 64322)2( 2 =++−=−y , so the graph contains the point ( – 2, 6).

The parabola looks like this

89


Note. It is not needed to always show dotted line for the line of symmetry. Final picture may look like this:

This graph may also be obtained from the graph of 22y x= by shifting three units to the left and four units up.

Notice that any quadratic function cbxaxy ++= 2 can be represented in the squared

form ( )2y a x h k= − + (and, therefore, its graph has the same shape as above). Actually, we already did this in the previous session when we discussed the method of the completing the square for quadratic equation. Let’s repeat this procedure one more time (with a small modification). We can use these steps.

1) On the right side of cbxaxy ++= 2 , factor out the coefficient a from the first two terms:

cxabxay +

+= 2 .

2) Divide coefficient ab by 2 and square it: 2

22

42 ab

ab

=

.

3) Inside parentheses (in step 1), add and subtract the above expression:

2

2222

4222

ab

abx

abxx

abx −

+⋅+=+ .

4) The first three terms on the right side of the above expression can be written as 2

2

+

abx , therefore

2

222

42 ab

abxx

abx −

+=+ .

5) Write the expression for y as

( ) 432 2 ++= xy

90


ca

ba

bxacxabxay +

−

+=+

+= 2

222

42

a

baca

bxaca

ba

bxa4

4242

2222 −+

+=+−

+= .

We’ve got the squared form ( )2y a x h k= − + of the parabola cbxaxy ++= 2 :

abac

abxay

44

2

22 −+

+= ,

where

2bha

= − , 24

4ac bk

a−

= .

Actually, to draw the graph of the parabola cbxaxy ++= 2 , you do not need to go through the above steps every time. Just memorize the most important formula for the x-coordinate of the vertex of parabola, which we denote as vx : This is the first coordinate h for the vertex (h, k). The second coordinate k is the value

244v

ac bk ya−

= = . This formula is more complicated and you do not need to memorize it.

This coordinate can be calculated by substitution the value of vx for x in the original parabola cbxaxy ++= 2 . Here are possible steps to graph the general parabola cbxaxy ++= 2 .

1) Identify coefficients a, b, and c.

2) Calculate the x-coordinate vx of the vertex of the parabola: a

bxv 2−= or by

competing the square. 3) Calculate the y-coordinate vy of the vertex by substituting vx in the original equation.

4) Follow the above steps for graphing the parabola ( )2y a x h k= + + , where ,v vh x k y= − = .

Example 11.2. Graph the parabola 582 2 −+−= xxy . Solution. 1) Identify the coefficients a, b, and c: a = –2, b = 8, c = –5.

2) Calculate the x-coordinate of the vertex: 2)2(2

82

=−⋅

−=−=a

bxv .

abxv 2

−=

91


3) Calculate the y-coordinate of the vertex by substitution 2=vx in the original equation: 352822 2 =−⋅+⋅−=vy . So, the vertex of the parabola has the coordinates (2, 3).

4) Draw the parabola according to the steps described for the squared form ( )2y a x h k= + + . In particular, parabola opens down (because a = –2 < 0) and has a

vertical line of symmetry that passes through the vertex (2, 3). Also (for more accuracy), we can calculate the values y(0) = –5, and y(1) = 1. Here is the picture.

Note. In order to plot the parabola more accurately, you may want to find the x- and y-intercepts. It is easy to find the y-intercept. Indeed, any point on the y-axis has the first coordinate zero, so just substitute x = 0 into the function cbxaxy ++= 2 , and you will find that the y-intercept is equal to c. To find x-intercepts, substitute y = 0 and solve the quadratic equation 2 0ax bx c+ + = . Consider in more details how parabola shows possible cases about the number of real solutions (roots) to the equation 02 =++ cbxax . In general, solving the equation

0)( =xf means to find all values of x for which the function )(xfy = takes the value of zero: y = 0. Geometrically, points (x, 0) lie on the x-axis. Therefore, roots of the equation

0)( =xf are x-coordinates of points of intersection of the graph of )(xfy = with the x-axis. Thus, geometrically solving the equation f(x) = 0 is equivalent to finding all x-intercepts of the graph of the function )(xfy = . In particular, to find real solutions to the quadratic equation 02 =++ cbxax , we need to find all x-intercepts of the parabola cbxaxy ++= 2 . We consider the case a > 0 (parabola opens up). The case a < 0 is similar. Obviously there are only three possible positions of the parabola with respect to x-axis:

1) The vertex of the parabola is located below x-axis. In this case there are two x-intercepts, so there are two roots of the quadratic equation.

2) The parabola touches the x-axis at one point (at the vertex), so there is only one root.

3) The parabola is located above x-axis. In this case, no x-intercepts, so no real roots.

582 2 −+−= xxy

92


Here are the corresponding pictures:

Two roots One root No roots

For example, parabola 2 4y x= − has two roots 2 and –2 (since y(2) = y(–2 ) = 0), parabola 2y x= has only one root 0, and parabola 2 4y x= + does not have roots at all (since the equation 2 4 0x + = does not have real solutions).

93


Exercises 11 In exercises 11.1 and 11.2,

1) Identify the coordinates of the vertex of given parabola. 2) Determine whether the parabola opens up or down.

11.1. a) 23( 4) 5y x= + −

b) 24( 2) 6y x= − − +

c) 27( 5) 4y x= − −

11.2. a) 25( 3) 7y x= − − +

b) 26( 7) 3y x= + −

c) 22( 1) 8y x= − + + In exercises 11.3 and 11.4, the given graphs are shifted graphs of the parabola 23y x= . Write the equations for the parabolas graphed below.

1) In squared form 2( )y a x h k= − + .

2) In general form 2y ax bx c= + + .

11.3. a) b)

94


11.4. a) b) In exercises 11.5 and 11.6, the given graphs are shifted graphs of the parabola 22y x= − . Write the equations for the parabolas graphed below.

1) In squared form 2( )y a x h k= − + .

2) In general form 2y ax bx c= + + .

11.5.

a) b)

95


11.6. a) b)

In exercises 11.7 and 11.8, for each of the quadratic functions

1) Find the vertex of parabola. 2) Find the y- intercept. 3) Find the x- intercepts if they exist. 4) Write the equation of the line of symmetry. 5) Graph the parabola. 6) Label the vertex and the x- and y- intercepts with coordinates.

11.7. a) 2 2 3y x x= − −

b) 2 4 5y x x= − − +

11.8. a) 2 2 8y x x= − − +

b) 2 6 5y x x= − +

96

Session 12: Distance Formula, Midpoint Formula, and Circles

Session 12

Distance Formula, Midpoint Formula, and Circles When we say that a point in the x-y-plane is given we mean that coordinates of the point are given. If A is a point in the plane and (x, y) are its coordinates in the coordinate system, then we will also denote this point by ( , )A x y .

Distance Formula Assume two points 1 1( , )A x y and 2 2( , )B x y are given. Consider the problem how to find the distance between them (i.e. to get a formula for their distance).

Recall that if we plot a point 0 0( , )C x y in the coordinate system, then 0x is the horizontal coordinate (along the x-axis), and 0y is the vertical coordinate (along the y-axis):

Now consider two points 1 1( , )A x y and 2 2( , )B x y :

We denote by ( , )d A B the distance between the points A and B (i.e. the length of the segment AB). Also, we will the notation AB , so ( , )d A B AB= . To find this distance, we draw a right triangle with the hypotenuse AB and horizontal and vertical legs:

y

x

•

0x

0y

0 0( , )C x y

x

y

1x

1y

1 1( , )A x y

2 2( , )B x y

2y

2x

x

y

2 1x x− 1y

2 1y y−

2x 1x

A

B 2y

C

97


We can see that 2 1AC x x= − and 2 1BC y y= − . By the Pythagorean Theorem,

( ) ( )2 22 2 222 1 2 1( , )d A B AB AC BC x x y y= = + = − + − .

By taking square root of both sides, we get the Distance Formula:

( ) ( )2 22 1 2 1( , )d A B x x y y= − + −

Here 1 1( , )x y and 2 2( , )x y are coordinates of the points A and B respectively.

Note. The distance formula will not change, if we switch (exchange) 1x and 2x , and/or

1y and 2y . The reason is that if a and b are two numbers, then 2 2( ) ( )a b b a− = − . Also, the distance formula is valid for points located in any quadrant (not only in the 1st quadrant).

Example 12.1. Calculate the distance between the points (2, 3)A − and ( 5, 7)B − − .

Solution. By distance formula we get:

( ) ( ) ( ) ( )2 2 2 2 2 22 ( 5) 3 ( 7) 2 5 3 7 7 4 49 16 65AB = − − + − − − = + + − + = + = + = .

Later, in Session 18 “Solving Oblique Triangles – Law of Cosines”, in Example 18.4 we will justify a method for checking whether a triangle with given sides a, b, and c is acute, obtuse or a right triangle. Here are the pictures of such triangles: Acute triangle Right triangle Obtuse triangle

The method is the following: Let c be the biggest side of the triangle. Calculate the value 222 cbaE −+= . If E > 0, then the triangle is acute. If E < 0, then the triangle is obtuse. If E = 0, then the triangle is right. Example 12.2. Consider a triangle ABC with vertices (2, 3), ( 5, 7)A B− − − and ( 2, 6)C − . Determine what kind of triangle it is: acute, obtuse, or a right triangle. Also, determine which angle in the triangle ABC is the biggest angle, and which angle is the smallest one.

Solution. First, we calculate squares of all three sides of triangle ABC (there is no need to calculate the sides themselves because the above expression for E contains squares of sides only). Length of the side AB we already calculated in Example 12.1, so 2 65.AB = Using the distance formula for the sides AC and BC we have:

98


( ) ( ) ( ) ( )2 2 2 22 2 ( 2) 3 6 2 2 9 16 81 97AC = − − + − − = + + − = + = ,

( ) ( ) ( ) ( )2 2 2 22 5 ( 2) 7 6 5 2 13 9 169 178BC = − − − + − − = − + + − = + = .

We see that side BC is the longest one. Now we can construct the above expression for E:

2 2 2 65 97 178 16E AB AC BC= + − = + − = − .

Since E is negative, we conclude that the triangle ABC is obtuse. In any triangle, the larger the side, the greater the opposite angle. In our case, the largest side is BC, and the smallest side is AB. Therefore, the largest angle is at A, and the smallest one is at C. Midpoint Formula Let A and B be two points. The midpoint formula gives the coordinates of the point C located in the middle of the line segment AB. To get these coordinates, consider first the simplest case when points A and B lie on the number line (i.e. on the horizontal x-axis):

Here point C is the midpoint of the segment AB. The location of any point on the number line is determined by its coordinate (a number). Let a, b, c be the coordinates of points A, B, C respectively. The distance between the points A and B (i.e. the length of the segment AB) is AB b a= − . It is easy to show that the coordinate c of the midpoint C is the

average of coordinates a and b: 2

a bc += . Indeed, with this coordinate, the distance

AC is half of the distance AB :

2

2 2 2 2ABa b a b a b aAC c a a+ + − −

= − = − = = = .

Now, let the points A, B and the midpoint C lie in the plane and have coordinates 1 1 2 2( , ), ( , )x y x y and ( , )m mx y respectively:

We see that mx is the midpoint of the segment 1 2[ , ]x x on the x-axis, and my is the midpoint of the segment 1 2[ , ]y y on the y-axis. Therefore, mx is the average of 1x and 2x , and my is the average of 1y and 2y . We thus obtained to the Midpoint Formula:

A C B x

x

y

A

1x 2x

1y

2y

B

C my

mx

99


The coordinates ( , )m mx y of the midpoint C of the line segment AB are averages of the corresponding coordinates of the endpoints 1 1( , )A x y and 2 2( , )B x y :

1 2 1 2,2 2m m

x x y yx y+ += =

Example 12.3. Calculate the coordinates of the midpoint of the line segment with endpoints ( 3, 4)− and ( 7,10)− . Solution. Let ( , )m mx y be coordinates of the midpoint. By the midpoint formula

3 ( 7) 10 52 2mx − + − −

= = = − and 4 10 14 72 2my +

= = = .

Answer: the midpoint has coordinates ( 5,7)− .

Example 12.4. Let (4, 7)A − be an endpoint of a line segment, and ( 6, 9)C − be its midpoint. Find the coordinates of the other endpoint of the line segment. Solution. Denote the other endpoint by ( , )B x y . We will use the midpoint formula with the given midpoint ( 6, 9)C − , so 6mx = − and 9my = . We have

46 12 4 16,2m

xx x x+= − = ⇒ − = + ⇒ = −

79 18 7 25.2m

yy y y− += = ⇒ = − + ⇒ =

Answer: endpoint B has coordinates ( 16, 25).− Circle

By definition, a circle is a set of points in the plane equidistant (having the same distance) to a fixed point in the plane. This fixed point is called the center of the circle, and the distance from any point on the circle to the center is called the radius. Equation of a circle can be easily derived directly from the distance formula. Let ( , )C a b be a center of a circle, and ( , )A x y be any point on the circle. If r is the radius of the circle, then, by definition, ( , ) .d A C r=

By distance formula, ( ) ( )2 2( , )d A C x a y b r= − + − = . Square both sides, and get

A r

C

x

y

100


Equation of the Circle:

( ) ( )2 2 2x a y b r− + − =

This equation is called the equation of circle in standard form. Here ( , )a b are coordinates of the center of the circle, and r is its radius. Example 12.5. Identify the center and the radius of the circle ( ) ( )2 23 5 15x y− + + = .

Solution. This equation is given in standard form, and we immediately get the answer: the center has coordinates (3, 5)− , and the radius is 15 .

Note. Notice that in the above example, the second coordinate of the center is 5− , not 5. This is because according to the equation of circle, we have 5 ( 5)y b y y− = + = − − , so

5b = − . Also, the radius is equal to 15 , but not 15, since number 15 is the square of the radius: 2 15r = . Equation of a circle may be given in a non-standard form. In this case, to identify the center and the radius of this circle, we represent the equation of circle in the standard form first. To achieve this, we can apply the completing the square technique. As a review, you may take a look at Session 10 “Completing the Square and the Quadratic Formula”. Example 12.6. Identify the center and radius of the circle given by the equation

2 2 8 10 32 0x y x y+ + − + = . Solution. We reorganize the terms and write the equation like this

( ) ( )2 28 10 32 0x x y y+ + − + = .

Now complete the square for both x and y. According to the procedure for completing the square that we described in Session 9, in each pair of parenthesis we divide coefficients for x and y by 2 and square them: 2 2(8 / 2) 16, ( 10 / 2) 25.= − = Then we add 16 and 25 to both sides of the equation:

( ) ( ) ( ) ( )2 22 28 16 10 25 32 16 25 4 5 32 41,x x y y x y+ + + − + + = + ⇒ + + − + =

( ) ( ) ( ) ( )2 2 2 24 5 41 32 4 5 9.x y x y+ + − = − ⇒ + + − = We obtained the equation of circle in the standard form. From here, the coordinates of the center are ( 4, 5)− and the radius is 3. Example 12.7. Let ( 4,3)A − and ( 6,9)B − be the endpoints of a diameter of a circle. Find the equation of the circle in the standard form. (Diameter of a circle is a line segment that passes through the center and whose endpoints lie on the circle).

101


Solution. First, let’s find the coordinates of the center ( , )C a b of the circle. The center is the midpoint of the diamenter AB. Using the Midpoint Formula, we have

4 ( 6) 10 52 2

a − + − −= = = − , 3 9 12 6

2 2b += = = .

So, the center has the coordinates ( 5, 6)− . Next, we will find the radius r of the circle. It is equal to the distance from the center of the circle to any point on the circle. To calculate the radius, we can take any of the given points: A or B. Let’s take the point

( 4,3)A − . Using the Distance Formula, we have

( ) ( ) ( )2 2 2 2 2 2( 5) ( 4) 6 3 5 4 3 ( 1) 9 1 9 10 10r r= − − − + − = − + + = − + = + = ⇒ = .

Now, using the coordinates ( 5, 6)− of the center of the circle and the square of its radius, which is 10, we can write the equation of the circle in the standard form:

( ) ( )2 2( 5) 6 10x y− − + − = , or ( ) ( )2 25 6 10x y+ + − = . Example 12.8. Graph the circle from example 12.6 and label four points on the circle. Solution. In example 12.6, we found that the center of the circle is ( 4, 5)− and its radius is 3. We use this information to graph the circle via the following steps:

1) Plot the center ( 4, 5)− .

2) From the center, draw dotted horizontal and vertical lines. 3) Along these lines, count 3 units (which is radius) starting from the center in all

four directions: up, down, left and right. Mark the four corresponding points as A, B, C, and D. These points are points on the circle.

4) Draw the circle through the points A, B, C, and D. The points A, B, C, D have coordinates ( 4, 8), ( 4, 2), ( 7, 5),A B C− − − and ( 1, 5).D −

y

x –1 –2 –3 –4 –5 –6 –7

1 2 3 4 5 6 7 8

0

A

B

D C

102


Exercises 12 In exercises 12.1 and 12.2, calculate the distance between the given points. 12.1. (5, 4) and ( 1, 2)− 12.2. ( 2, 5)− and ( 3, 12)− In exercises 12.3 and 12.4, three vertices A, B and C of the triangle ABC are given.

1) Determine what kind of triangle it is: acute, obtuse, or a right triangle. 2) Determine which angle in the triangle ABC is the biggest angle, and which angle is the

smallest one. (You may want to review Example 12.2) 12.3. ( 2, 3), (5, 2), (6, 5)A B C− − − 12.4. ( 2, 4), ( 7, 1), (4, 2)A B C− − − − In exercises 12.5 and 12.6, calculate the coordinates of the midpoint of the line segment with given endpoints. 12.5. (3, 4)− and ( 5, 6)− − 12.6. (4, 5)− and (2, 7) In exercises 12.7 and 12.8, C is the midpoint of the line segment AB. The coordinates of points A and C are given. Find the coordinates of the point B. 12.7. ( 9, 7), ( 4, 5)A C− − 12.8. (5, 8), ( 1, 3)A C − In exercises 12.9 and 12.10, identify the center and the radius of the given circles. 12.9. a) ( ) ( )2 22 4 36x y+ + + =

b) ( ) ( )2 25 2 20x y+ + − =

12.10. a) ( ) ( )2 26 3 49x y− + − =

b) ( ) ( )2 27 8 50x y− + + =

In exercises 12.11 and 12.12, points A and B are given endpoints of a diameter of a circle. Find the equation of the circle in standard form ( ) ( )2 2 2x a y b r− + − = . 12.11. (2, 8), ( 8, 16)A B − − 12.12. (20, 25), (4, 5)A B − In exercises 12.13 and 12.14, equations of circles are given. For each circle

1) Find the coordinates of the center and the radius. 2) Label four endpoints of vertical and horizontal diameters with their coordinates. 3) Graph the circle.

12.13. a) 2 2 4 2 11 0x y x y+ + − − =

b) 2 2 10 8 32 0x y x y+ − − + =

12.14. a) 2 2 8 2 8 0x y x y+ − + + =

b) 2 2 6 4 12 0x y x y+ + + − =

103

Session 13: Nonlinear Systems of Equations in Two Variables

Session 13

Nonlinear Systems of Equations in Two Variables We now consider examples of systems of two equations with two variables in which one or both equations are not linear. We also give a geometrical interpretation of that solutions.

Example 13.1. Solve the system of equations

−+=

=+

846

2 xxyyx

Solution. Notice that the first equation is linear, while the second is not (since it contains 2x ). We can easily solve the first (linear) equation for x or y and substitute this

expression into the second (nonlinear) equation. As a result, the second equation will contain only one variable. This method is called the substitution method. Initially, it does not matter for which variable we solve the first equation: for x or for y. However, it does matter for the second equation. If we solve the first equation for x, then we need to substitute this expression in the second equation for both 2x and x, which is a bit involved. It is more suitable to solve the first equation for y. In this case we substitute this expression into the second equation for y, and we will not need to square any expression. So, solve the first equation for y: xy −= 6 . Substitute this expression into the second equation, and get a quadratic equation in the variable x:

846 2 −+=− xxx . Solve this equation:

26 4 8x x x− = + − ⇒ 2 4 8 6 0x x x+ − − + = ⇒ 2 5 14 0x x+ − = ( 2)( 7) 0x x⇒ − + = ⇒ x = 2 and x = –7.

Now, using the expression xy −= 6 , we can find the corresponding values of y: if 2=x , then y = 6 – 2 = 4, and if x = –7, then y = 6 – (–7) = 13.

Final answer: the system has two solutions: x = 2, y = 4, and x = –7, y = 13. Or, as a solution set, { })13,7(),4,2( − .

Note. We can interpret the above solutions geometrically using graphs of the given equations. The graph of the first equation is a straight line, and the graph of the second equations is a parabola. Solutions of the system give the points of intersection of these two graphs. According to the final answer, the straight line and the parabola intersect each other at two points with the coordinates (2, 4) and (–7, 13). Here is the corresponding picture:

104


Example 13.2. Solve the system of equations

−+=

+−=

46143

2

2

xxyxxy

.

Solution. In this example both equations are nonlinear and both are solved for y. We can use different methods to solve this system. For example, we can equate the right sides of both equations. We can also eliminate y by subtracting one of the two equations from the other. Using this method, we subtract the first equation from the second and get

( )2 2 2 26 4 3 14 6 4 3 14 9 18y y x x x x x x x x x− = + − − − + = + − − + − = −

0 9 18 9 18 2x x x⇒ = − ⇒ = ⇒ = .

To find y, we can substitute x = 2 in either equation of the system. Substituting it into the first equation, we get 121464142322 =+−=+⋅−=y .

Final answer: the system has one solution x = 2, y = 12, or as a pair (2, 12), or as a solution set { }(2, 12) .

Note. As in example 13.1, we can interpret the above solution geometrically. The graphs of both equations of the system are parabolas. According to the final answer, these parabolas intersect each other only at one point (2, 12):

(2, 4) (– 7, 13)

(2, 12)

2 3 14y x x= − +

2 6 4y x x= + −

105


Example 13.3. Solve the system of equations 2 2

2 2

3 2 142 5 13x yx y

+ =

+ =.

Solution. Notice that in this system both variables, x and y, are in the second power only. We may temporary use new variables u and v: 2 2,u x v y= = . Then, in terms of u and v, we have a linear system

3 2 144 3 13u vu v+ =

+ = .

Solve this system by the elimination method:

3 2 14 6 4 282 5 13 6 15 39u v u vu v u v+ = − − = −

⇒ + = + = .

Add the last two equations to eliminate u and solve for v: 11 11 1v v= ⇒ = . Substitute this value into the first equation of the system, and solve for u:

3 2 1 14, 3 2 14, 3 12, 4u u u u+ ⋅ = + = = = .

So, u = 4 and v = 1. Now, we need to return from u and v to the original variables x and y. We have 2 4x u= = . From here, 4 2x = ± = ± . So, we obtained two values for x: 2 and –2. Similarly, 2 1 1 1y v y= = ⇒ = ± = ± . We have two values of y: 1 and –1.

From this point we need to be very careful to write the final answer in the correct way. Any solution of the given system is a pair (x, y). Therefore, we need to combine each value of x with each value of y. As a result, the original system has four solutions:

{(2, 1), (2, –1), (–2, 1), (–2,–1)}.

Note. It can be shown that graphs of the equations in the given system are ellipses (“stretched” circles centered at the origin (0, 0)). The answer to the problem tells us that these two ellipses intersect each other at the above four points:

–2 3

(– 2, 1)

(2, –1) (– 2, –1)

(2, 1)

2 23 2 14x y+ =

2 22 5 13x y+ =

106


Example 13.4. Solve the system of equations 2 22 15

y xx y

=

+ =.

Solution. Here both equations are non-linear. Square the first equation: 2y x= . Substitute this expression into the second equation and get a quadratic equation in x:

2 22 15 2 15 0 ( 3)( 5) 0x x x x x x+ = ⇒ + − = ⇒ − + = .

From here we get two solutions of the quadratic equation: x = 3 and x = – 5. Now we can use the first equation to find the corresponding values of y.

For x = 3, 3y = . So, one solution is the pair ( )3, 3 . For x = – 5, 5y = − . In this

session, we consider only real numbers. Because 5− is not a real number, we reject it. Finall answer: the original system has only one solution ( )3, 3 .

Note. Geometrically, the graph of the first equation is the curve, which has the shape of the half of parabola that “lies on the side”: it is going not along y-axis, but along x-axis, and its graph is located above x-axis. The second equation is the ellipse with the center in origin. Both curves intersect each other only at one point:

Example 13. 5. The area of a rectangular region is 96 square feet, and the perimeter is 40 feet. Find the dimensions of the region (i.e. find its length and width). Solution. As for most word problems, we will solve it in two steps: set up the equations for the values in question, and solve these equations. Also, check that the solution(s) make sense. 1) Let x be the length of the rectangle, and y be its width. Then 96=xy (area of the

rectangle), and 2x + 2y = 40 (perimeter of the rectangle). We get the following system of equations:

=+=

402296

yxxy

2) We can simplify the second equation by dividing both sides by 2:

( )3, 3

107


=+=

2096

yxxy

.

Here the second equation is a linear one, and we can easily solve it for x or y. Let’s solve it for y: xy −= 20 . Substituting this expression into the first equation, we obtain x(20 – x) = 96, or 9620 2 =− xx . This is a quadratic equation that can be rewritten in the standard form 096202 =+− xx . We can solve it by factoring:

( )8)( 12 0 8x x x− − = ⇒ = and 12x = .

We can get the corresponding value of y from the expression xy −= 20 . If x = 8, then 12820 =−=y . If 12=x , then 81220 =−=y .

Note. It looks like we have found two solutions: x = 8, y = 12, and x = 12, y = 8. These are the solutions of the system of equations. However, for the original problem that asks about the dimensions of the rectangle, these two solutions simply mean that one side of the rectangle is 8 feet, and the other is 12 feet. Assuming that the length is greater than the width, we come up to a unique solution.

Final answer: the length of the rectangle is 12 feet, and the width is 8 feet.

108


Exercises 13

In exercises 13.1 and 13.2, solve the systems of equations (find all real solutions).

13.1. a) 2

3 12 5

x yx y− =

+ =

b) 2

2

5 23 26

x y yx y y

= + −

= − −

c) 2 2

05 36

x yx y

+ =

− =

d) 2 2

2 33 2

x yx y− =

− = −

e) 2 2

2 2

2 345 3 93

x yx y

+ =

+ =

f) 2 2

2 2

2 74 3 31

x yx y

− = −

+ =

13.2. a)

2

4 23 19

x yx y

+ =

+ =

b) 2

2

6 23 16

x y yx y y

= + −

= + +

c) 2 2

03 4

x yx y

− =

− =

d) 2 2

3 28 4

x yx y+ =

− =

e) 2 2

2 2

5 3 283 2 44

x yx y

− = −

+ =

f) 2 2

2 2

3 375 2 70

x yx y

+ =

+ =

In exercises 13.3 and 13.4, A represents the area of a rectangular region, and P represents its perimeter. Find the dimensions of the region (i.e. find its length and width, assuming that the width does not exceed the length). 13.3. A = 12 m2, P = 14 m. 13.4. A = 30 yd2, P = 22 yd .

Challenge Problems

13.5. Consider the system of equations 2 2 2

x y ax y b+ =

+ =

, where 0a > and 0b > .


1) If 2a b> , then the system does not have solutions.

2) If 2a b= , then the system has one solution.

3) If 2a b< , then the system has two solutions. Interpret these results geometrically.

109


13.6. Consider the system of equations ( )2 2

0x yx a b y ab

− =

+ − =.

Prove the following statements: 1) If 0a = and 0b = , then the system has one solution ( )0, 0 .

2) If 0a > and 0b < , then the system does not have solutions.

3) If 0a > and 0b ≥ , then the system has one solution ( ),b b .

4) If 0a ≤ and 0b < , then the system has one solution ( ),a a− − .

5) If 0a < and 0b > , then the system has two solutions ( ),a a− − and ( ),b b .

13.7. Solve the system of equations 2 ( )

ax y bx cy d d ac bc− =

+ = + −

.

Hint: Check that the quadratic equation 2 ( ) ( )x c ax b d d ac bc+ − = + − has roots d and –d – ac.

110

Part II

Trigonometric Functions

111

Session 14: Geometric and Trigonometric Angles

Session 14

Geometric and Trigonometric Angles Historically, trigonometry studies the relationships between angles and sides of triangles. In Greek, the word “Trigonometry” literally means “Triangle-Measurement”. It is important to understand that in geometry and trigonometry we treat angles in different ways.

In geometry, an angle is simply a figure, created by two rays, coming from the same point. Also, we assign the measure to an angle as some positive number. A common measure is the degree measure. If you cut a round pizza-pie (theoretically) into 360 equal slices, the angle in one slice is of one degree: 1 .

In trigonometry, we extend the meaning of an angle by assigning to it the “direction of rotation” and, as a result, the sign of its measure. That means that we assign to angles not only positive measure, but also negative. We can do this in the following way. Consider “geometric” angle Let’s call one of its sides initial side, and the other – terminal side. Let’s say, the horizontal side is the initial, and the slant side is the terminal. We can treat this angle as a result of rotation of the terminal side when it starts from the position of initial side and then rotates to its current position. To rotate, we have two directions: clockwise and counterclockwise. We can mark these two directions of rotation by arrows: It was an agreement to assign to an angle a positive measure if the direction of rotation is counterclockwise, and assign a negative measure if the direction of rotation is clockwise. On the left picture above, the angle is positive, and on the right – negative. As you can see, taking one “geometric” angle (two rays, coming from the same point), we can consider two “trigonometric” angles: one positive and another negative depending

Initial side

Terminal side

Counterclockwise rotation: angle is positive.

Clockwise rotation: angle is negative.

112


on direction in which we rotate the terminal side. Even more, we can assign to a given “geometric” angle infinite many “trigonometric” angles making multiple full rotations of terminal side in either direction. All such “trigonometric” angles have the same “geometric” angle and they are called coterminal angles. On two pictures above, the angles are coterminal.

Example 14.1. Consider the angle of 40 : Describe all coterminal angles for this angle.

Solution. If we make one full rotation (rotation by 360 ) of the terminal side in either direction, the terminal side returns to its original position and we obtain coterminal angle (i.e. the same “geometric” angle). We get the same result, if instead of one, we make n full rotations (i.e. rotations by 360 n⋅ ). All such angles are coterminal to 40 angle and their values are described by the parametric formula 40 360 n+ ⋅ , where parameter n is any integer (positive, negative, or zero). We can write that 0, 1, 2,...n = ± ± . For positive n, we get positive values of the angle, and for negative n – negative values. For example, if we put 1=n and 1−=n , we get two specific coterminal angles: 40 360 400+ = and 40 360 320− = − . Two Special Right Triangles and Three Special Angles

In trigonometry, we often use the following two right triangles: one is a half of an equilateral triangle, and another is a half of a square:

In the triangle ABC on the left picture, the acute angles are of 30 and 60 . We will call such triangle a 6030 − triangle. In the triangle ABC on the right picture, both acute angles are of 45 . We will call such triangle a 4545 − triangle. Both triangles:

40 Initial side

Terminal side

A

B

C

b

a

c

A

B

C

a

b

c

ABC∆ is a half of an equilateral triangle

ABC∆ is a half of a square

113


6030 − and 4545 − , are called special right triangles, and angles 30 , 45 and 60 are called special angles.

Let’s consider special triangles in more details. For both, we will use the Pythagorean Theorem that states that for any right triangle with the hypotenuse c and legs a and b, the following equation is true: 222 cba =+ . We will also use this theorem in the forms:

2 2 2 2 2 2, ,c a b a c b b c a= + = − = − .

o o30 60− Triangle Let’s draw this triangle like this Recall that side c is the side of drawn above equilateral triangle, so side a is the half of

side c: 2ca = or c = 2a. Try to remember this fact:

Example 14.2. Consider 6030 − triangle with legs a, b and hypotenuse c (see picture above). Solve the following problems.

1) a = 7. Find b and c. 2) b = 5. Find a and c. 3) c = 10. Find a and b.

Solution. In all problems, side a is opposite to 30 angle. Therefore, c = 2a.

1) 2 2 7 14c a= = × = . By the Pythagorean Theorem 2 2 2 214 7 196 49 147 49 3 7 3b c a= − = − = − = = ⋅ = .

2) By the Pythagorean Theorem 2 2 2c a b= + and c = 2a. Therefore,

( )2

22 2 2 2 2 2 2 2 2 252 4 33 3bc a a b a a b a b a= = + ⇒ = + ⇒ = ⇒ = = .

25 5 5 3 10 3, 23 3 33

a c a= = = = = .

30

60

b

a c

In any 6030 − triangle, the leg opposite to 30 is the half of the hypotenuse (or the hypotenuse is twice as this leg).

114


3) Again, c = 2a, so 10 5

2 2ca = = = . By the Pythagorean Theorem

2 2 2 210 5 100 25 75 25 3 5 3b c a= − = − = − = = ⋅ = .

Let’s describe the connection between sides of 6030 − triangle in general form. Let a be a side, opposite to 30 . Then the hypotenuse c = 2a. Another side b which is opposite to 60 , can be calculated by the Pythagorean Theorem:

( )22 2 2 2 2 22 4 3 3b c a a a a a a a= − = − = − = = . We get the following picture

If you memorize this picture (or quickly get it), you can solve problems like in example 15.2 faster.

o o45 − 45 Triangle This triangle looks like this Both sides a and b are sides of the square above, therefore, they are equal: a = b. Try to remember this fact: Example 14.3. Consider 4545 − triangle with legs a, b and hypotenuse c (see picture above). Solve the following problems.

1) a = 5. Find b and c. 2) b = 7. Find a and c. 3) c = 10. Find a and b.

a

b

c

45

45

30

60

3a

a 2a

In any 4545 − triangle both legs are equal.

115


Solution. In all problems a and b are two legs, so they are equal: a = b. Therefore, problems 1) and 2) actually the same (just numbers are different).

1) a = b = 5. By the Pythagorean Theorem

2 2 25 25 50 25 2 5 2c a b= + = + = = ⋅ = .

2) a = b = 7, 2 2 2 2 27 7 7 2 7 2c a b= + = + = ⋅ = .

3) By the Pythagorean Theorem and using that a = b,

2 2 2 2 2 2 2 2, , 2 100, 50, 50 25 2 5 2a b c a a c a a a b+ = + = = = = = = ⋅ = .

Similar to 6030 − triangle, let’s describe the connection between sides of 4545 − triangle in general form. Let a be a side, opposite to one of the 45 angle. Then the other side b which is opposite to another 45 angle, is the same: b = a. The hypotenuse c can be calculated by the Pythagorean Theorem:

2 2 2 2 22 2c a b a a a a= + = + = = . We get the following picture

a

a

2a

45

45

116


Exercises 14 In exercises 14.1 and 14.2, determine all the coterminal angles for angle θ . Also, indicate two positive and two negative particular coterminal angles (answers may vary). 14.1.

a) 50θ = .

b) 70θ = − .

14.2. a) 27θ = . b) 35θ = − .

In exercises 14.3 and 14.4, 30 60− triangle is given. In it, A, B and C are angles, and a, b, and c are sides, which are opposite to corresponding angles. 30A∠ = , 90C∠ = . Solve the given problems. 14.3.

a) a = 6. Find b and c. b) b = 3. Find a and c. c) c = 8. Find a and b.

14.4. a) a = 8. Find b and c. b) b = 9. Find a and c. c) c = 4. Find a and b.

In exercises 14.5 and 14.6, 45 45− triangle is given. In it, A, B and C are angles, and a, b, and c are sides, which are opposite to corresponding angles. 45A∠ = , 90C∠ = . Solve the given problems. 14.5.

a) a = 4. Find b and c. b) b = 8. Find a and c. c) c = 9. Find a and b.

14.6. a) a = 3. Find b and c. b) b = 6. Find a and c. c) c = 7. Find a and b.

Challenge Problems 14.7. In 6030 − triangle, the hypotenuse is c. Express in terms of c two other sides. 14.8. In 6030 − triangle, the side opposite to 60 angle is b. Express in terms of b two

other sides. 14.9. In 45 45− triangle, the hypotenuse is c. Express in terms of c two other sides.

117

Session 15: Trigonometric Functions for Acute Angles

Session 15

Trigonometric Functions for Acute Angles Definition of six trigonometric functions Consider the following “giraffe” problem: “A giraffe’s shadow is 8 meters. How tall is the giraffe if the sun is 28 to the horizon?” Trigonometric functions that we introduce here, allow us to solve this and many more problems that involve angles and sides of triangles. We will solve the above problem in the example 15.1 below. To approach such problems, let’s start with definition of trigonometric functions for acute angles. Consider an acute angle θ :

Trigonometric functions (in short trig functions) take this angle as its argument (as input) and assign some numerical values to it (output values). You will see shortly what exactly these values are. Because angle θ is acute, we can always construct a right triangle with this angle: By proportionality properties of similar triangles, the ratios of sides of this triangle do not depend on the size of the triangle; instead, they depend only on the value of the angle θ . In other words, if we take two right triangles with the same angle θ , but different sizes, then the ratios of the corresponding sides remain the same, since the triangles are similar. Trigonometric functions are exactly these ratios.

It is easy to see that there are only six possible ratios of the sides in a triangle. Here are all of them: a/c, b/c, a/b, b/a, c/b, c/a. So, there are exactly six trigonometric functions. Each of them has its own name and notation. The following table defines all six trig functions for angle θ .

θ

θ

c

b

a

118


Function Name Function Notation Definition

sine θsin ca

cosine θcos c

b

tangent θtan ba

cotangent θcot a

b

secant θsec bc

cosecant θcsc a

c

It may seem difficult to memorize all these functions. A simple advice (but perhaps not so easy to follow) is just to memorize them as you would the multiplication table.

From the above six trig functions, the first three are the most frequently used: sine, cosine, and tangent. They are called basic trig functions. The other three are reciprocals to basics: cotangent is reciprocal to tangent, secant is reciprocal to cosine, and cosecant is reciprocal to sine:

1 1 1cot , sec , csc .tan cos sin

θ θ θθ θ θ

= = =

Some people like the following mnemonic device SohCahToa for remembering the definition of the basic trig functions. It works like this. In the above right triangle, we can treat legs a and b as opposite and adjacent to the angle θ :

Now, the definition of sine, cosine, and tangent can be reformulated as ypotenuseOpposite/Hsin =θ

ypotenuseAdjacent/Hcos =θ

djacentOpposite/Atan =θ

The first three letters of the word SohCahToa mean: Sine is the ratio of Opposite leg to Hypotenuse, so we get Soh, and so on.

θ

Hypotenuse

Adjacent

Opp

osite

119


Trig Functions for Special Angles In the previous session we have introduced three special angles 30 , 45 and 60 as angles in special right triangles 6030 − and 4545 − . Here we calculate the basic trig functions sine, cosine and tangent for these angles. Because trig functions do not depend on the size of a triangle, for calculations, we can choose any value for one of the sides. Let’s select the value of 1 for the shortest leg of

6030 − triangle and for both legs of 4545 − triangle. Recall that in 6030 − triangle, hypotenuse is twice as the shortest leg (this leg is opposite to 30 angle), so the

hypotenuse is 2. Then by Pythagorean Theorem the other leg is 2 22 1 3− = . For 4545 − triangle, hypotenuse is 2 21 1 2+ = . We can draw the following two

pictures

Now we use the definition of basic trig functions.

30 angle: opposite side is 1, adjacent side is 3 , and hypotenuse is 2. Therefore,

1 3 1 3sin 30 , cos30 , tan 30 .2 2 33

= = = =

60 angle: opposite side is 3 , adjacent side is 1, and hypotenuse is 2. Therefore,

3 1sin 60 , cos 60 , tan 60 3.2 2

= = =

45 angle: opposite side is 1, adjacent side is 1, and hypotenuse is 2 . Therefore,

1 2 1 2sin 45 , cos 45 , tan 45 1.2 22 2

= = = = =

We summarize these results in the following table

Angle θ 30 45 60

θsin 21

22

23

θcos 23

22

21

θtan 33

1 3

1 30

60

3

2 2 1

1

45

45

120


To restore the values of θsin for the above special values and even for 0θ = and 90θ = (we will define these values in the next session), the following expression can be

used: 2n . Just put n = 0, 1, 2, 3, and 4 according to the table

n 0 1 2 3 4 Angle θ 0 30 45 60 90

sin2nθ =

0 02

= 1 1

2 2=

22

23

4 1

2=

Working with arbitrary acute angles The practical application of trigonometry is essentially based on the following general principal: it is easier to measure angles than distances. To measure angles, there is an optical device which is called the clinometer. Schematically, it looks like this:

The angle from the horizontal line going up is called the angle of elevation. In the picture above, θ is the angle of elevation that can be measured by clinometer. In similar way, the angle from the horizontal line going down is called the angle of depression: To find the values of basic trig functions for arbitrary angles, we can use buttons sin, cos and tan on scientific or graphing calculator.

Example 15.1. Let’s solve the “giraffe” problem, stated at the beginning of this session. We can draw corresponding picture like this

θ

θ

angle of elevation angle of depression

121


For the 28 angle, giraffe is the opposite side, and the giraffe’s shadow is the adjacent side. A suitable trig function is tangent (ratio of the opposite side to adjacent). Let’s

denote giraffe’s shadow by s and giraffe’s height by g. We have tan 28 gs

= . From here,

tan 28 8 0.5317 4.25g s= = ⋅ = m. Example 15.2. Nick launched a kite on a 120-m thread. The angle of elevation of the thread is 37 . At what altitude is the kite flying?

Solution. Here is the corresponding picture

Let’s denote the length of the thread by t. This is the hypotenuse and t = 120. The problem is to find height h which is opposite to the 37 angle. A suitable trig function is

sine (ratio of the opposite side to hypotenuse). We have sin 37 ht

= . From here,

sin 37 120 0.6018 72.22h t= = ⋅ = m. Example 15.3. A ladder is leaning against the wall such that the angle of depression of the top of the ladder is 56 . What is the length of the ladder if the distance from its lower end to the wall is 2 m? Solution. We draw two dotted line segments: one horizontal and one vertical. They have the same lengths as those to which they are parallel, in particular, horizontal line segment is 2 m. In the right triangle formed by the ladder and dotted lines, the ladder is the

37

t h

56

wall

2 m

ladder

28 Shadow = 8 m

122


hypotenuse, and the top (horizontal) dotted line is the side adjacent to 56 angle. Let’s denote the length of this side by d and the ladder’s length by l. We have d = 2 m. The problem is to find l. A suitable trig function is cosine (ratio of the adjacent side to

hypotenuse). We have cos56 dl

= . From here,

2 3.58cos56 0.559

dl = = =

m.

Trig functions allow also to find angles in right triangles when info about sides is known. To solve such problems, first identify, similar to previous examples, which trig function relates to given problem and find the value of this function. Then, to find the angle you can use buttons 1 1sin , cos− − and 1tan− on calculator. These buttons calculate the values of so-called inverse trigonometric functions. These functions restore angles from the values of corresponding trig functions. We will say more about inverse trig functions in sessions 18 and 19. Example 15.4. A ship is 160 m away from the center of a horizontal barrier that measures 200 m from end to end. What is the minimum angle that the ship must be turned to avoid hitting the barrier?

Solution. The problem is to find angle θ . Let’s denote half of the length of barrier as a. We have a = 200/2 = 100 m. This is the side of right triangle on the picture and it is opposite to angle θ . Another side is the distance from the ship to the barrier. This side is adjacent to angle θ . We denote it as d. It is given that d = 160 m. Appropriate trig function here is tangent (ratio of the opposite side to adjacent).

( )1100tan 0.625 tan 0.625 32160

ad

θ θ −= = = ⇒ = = .

Example 15.5. An airplane is flying at an altitude of 2.5 miles and is preparing for landing. It is 8.6 miles from the runway. Find the angle of depression that the airplane must make to land safely.

Solution. The problem is to find the angle θ . We denote the path (distance) of the

runway

θ 2.5 mi 8.6 mi

Bar

rier

200

m

160 m θ

123


airplane to the runway as d, and altitude as h. It is given that d = 8.6 mi and h = 2.5 mi. In the drawn right triangle (with dotted lines), d is hypotenuse, and h is opposite side for angle θ . Appropriate trig function here is sine (ratio of the opposite side to hypotenuse).

( )12.5sin 0.29 sin 0.29 178.6

hd

θ θ −= = = ⇒ = = .

Example 15.6. Lillian wants to shingle her roof. The roofer asked her for the angle of elevation of the roof to make sure he can climb the roof safely. Help Lillian to calculate the angle according to the following picture:

Solution. The problem is to find angle θ . Let’s denote the marked horizontal line segment (half of the width of the house) as a, and slant line segment (the width of the roof) as s. We have a =26/2 = 13 ft and s = 14.5 ft. In the triangle on the right side of the picture, s is hypotenuse and a is the side adjacent to angle θ . Appropriate trig function here is cosine (ratio of the adjacent side to the hypotenuse).

( )113cos 0.9 cos 0.9 2614.5

as

θ θ −= = = ⇒ = = .

It looks like it is safe for the roofer to climb the roof.

θ

a

14.5 ft

26 ft

124


Exercises 15 In problems 15.1 and 15.2, find sine, cosine and tangent of angles A and B. 15.1. 15.2 15.3. Find the exact values of secant and cosecant of angles 30 , 45 and 60 . 15.4Find the exact value of cotangent of angles 30 , 45 and 60 . In problems 15.5 and 15.6, determine which angle is the angle of elevation: A or B. 15.5. 15.6. In problems 15.7 and 15.8, determine which angle is the angle of depression: A or B. 15.7. 15.8.

In the problems below, round answers to the nearest tenth.

15.9. The angle of depression from the top of an apartment building to the base of a fountain in a nearby park is 70 . If the building is 80 ft tall, how far away is the fountain?

15.10. Allison is looking at the top of a tall building. Her eyes are 5 feet above the ground. The angle of elevation is 75 and she is 15 feet from the building. How tall is the building?

15.11. A 20 foot ladder rests against a wall. Its angle of elevation from the ground is 55 . How far from the wall is the base of the ladder?

15.12. Nick needs to reach a top window of the house using a ladder. He wants to put a ladder 1.5 meters from the wall. At this point, he measured that the angle of elevation to the window is 53 . How long does the ladder have to be?

15.13. Ben is flying a kite and realizes that 260 feet of string are out. The angle of elevation of the kite is 40 . How high is kite above the ground?

A B A B

A B A

B

5

4

3 A

B

5 A

B

13

12

125


15.14. Lillian is swimming in the sea and notices a coral reef at the sea bottom. The angle of depression is 37 and the depth of the sea here is 50 feet. How far is she from the reef?

15.15. Over 6000 feet (horizontal), a road rises 330 feet (vertical). What is the angle of elevation?

15.16. Suppose a tree 15 m in height casts a shadow of length 27 m. What is the angle of elevation from the end of the shadow to the top of the tree?

15.17. A boat is sailing and spots a big shell 18 feet below the water. A diver jumps from the boat and swims 25 feet to reach the shell. What is the angle of depression from the boat to the shell?

15.18. A ladder leans against a wall. The foot of the ladder is 5.4 feet from the wall. The ladder is 15 feet long. Find the angle the ladder makes with the wall.

15.19. A vertical pole stands on the ground and has a support wire that runs from its top to the ground. The support is 50 feet long and anchored 22 feet from the base of the pole. Find the angle of elevation from the anchor point to the top of the pole.

15.20. Eli is putting up an antenna at the flat roof of a house. At its top, he attached a 50 ft guy wire and anchored it on the roof. Antenna is 30 ft long. What angle does the guy wire form with the antenna?

Challenge Problems

15.21. At a river shore (right near the water) a tree is standing. On the opposite side of the river, across the tree, Margaret is standing at the distance of 10 m from the water. She wants to determine the width of the river. She found that the angle of elevation to the top of the tree is 32 . Then Margaret walked right to the water and found that the angle of elevation now became 43 . What is the width of the river?

15.22. Esther and Nick stand at points E and N, 2 m apart in a dark room with a large mirror. Esther stands 2 m from the mirror, and Nick stands 1 m. At what angle EBA should Esther shine a flashlight on mirror so that the reflected light directly strikes Nick?

Note. According to the law of reflection, EBA NBC∠ = ∠ .

Mirror

E

N 2

2

1

A B C

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Session 16: Trigonometric Functions for Arbitrary Angles

Session 16

Trigonometric Functions for Arbitrary Angles Unit Circle In previous session we defined trig functions for acute angles: we constructed right triangle with given angle, and defined trig functions as ratios of sides in this triangle. This approach cannot be used for angles that are not acute like obtuse or negative angles: there are no right triangles with such angles. Nevertheless, it is possible to define trig functions for arbitrary angles. To do this we will use a special tool that allows to reformulate definition of trig function of acute angles in such a way that a new definition can be used for arbitrary angles. This tool is called the unit circle in the system of coordinates. This is just a circle with the radius of 1 and the center at the origin: In this figure we will draw angles in the standard position. It means that their vertices are at the origin, and the initial sides goes along the positive part of the x-axis. Here is an example of such angle θ in the 1st quadrant (i.e. acute angle): Angle θ is uniquely defined (up to coterminal angles) by the point A on the circle at which terminal side intersects the circle. We will call point A corresponding to the angle θ . Let (a, b) be coordinates of the point A (we also use the notation A(a, b) for point A):

y

x 1 0

x 1 0 θ

A . y

127


Notice that 0A = 1 (radius of the unit circle). Then from the right triangle 0AB, we have

bbA

AB===

10sinθ , aa

AB

===10

0cosθ .

We see that for acute angles, sine and cosine are the coordinates of the corresponding points on the unit circle: sine is the second coordinate (y-coordinate), and cosine is the first coordinate (x-coordinate). We’ve got the reformulation (i.e. a new definition) of sine and cosine for acute angles: they are the coordinates of points on the unit circle. We can use this reformulation as a general definition for arbitrary angles. Definition. Let θ be an arbitrary angle in standard position, and A(a, b) be the corresponding point on unit circle. Then, by definition,

sinsin , cos , tancos

bb aa

θθ θ θθ

= = = = .

Note. To remember which of the trig functions – sine or cosine – is the first coordinate, and which one is the second, you may use the alphabetical order of the first letters in the words sine and cosine (c is before s, so cosine is the first coordinate, and sine is the second). Other three trig functions can be defined as reciprocals to the basics:

θ

θtan

1cot = , θ

θcos

1sec = , θ

θsin

1csc = .

Because sine and cosine are coordinates, trig functions may take both positive and negative values depending on the quadrant in which angle θ lies. The following figures show the signs of basic trig functions.

+

Signs of sine

–

+

–

+

Signs of cosine

+

–

–

+

Signs of tangent

–

–

+

x 1 0

θ

A . B

y

sinb θ=

cosa θ=

128


Note. The following phrase may help you to remember which of these functions (sine cosine or tangent) is positive in each quadrant: “All Students Take Calculus”. This phrase hints that in the first quadrant all three are positive, in the second – only sine, in the third – only tangent, and in the fourth – only cosine.

Example 16.1. Calculate the basic trig functions for the quadrant angles of 0 , 90 , 180 , 270 , and 360 .

Solution. Here is the picture for quadrant angles:

1) For 0 and 360 angles the corresponding point on unit circle has coordinates (1, 0). Therefore,

0360sin0sin == , 1360cos0cos == , 0360tan0tan == .

2) For 90 angle the corresponding point has coordinates (0, 1) . Therefore,

190sin = , 090cos = . By definition, θθθ

cossintan = . Because 090cos = ,

90tan is undefined (we can not divide by zero).

3) For 180 angle the corresponding point has coordinates ( 1, 0)− . Therefore, 0180sin = , 1180cos −= , 0180tan = .

4) For 270 angle the corresponding point has coordinates (0, 1)− . Therefore, 1270sin −= , 0270cos = , 270tan is undefined.

We summarize the results of example 17.1 in the following table

Angle θ 0 90 180 270 360 θsin 0 1 0 –1 0 θcos 1 0 –1 0 1 θtan 0 undefined 0 undefined 0

We see that the maximal and minimal values of sine and cosine of the quadrant angles are 1 and – 1 respectively. For all other angles sine and cosine are between –1 and 1. In general, for any angle θ sin 1, cos 1θ θ≤ ≤ . There is no restriction for tangent.

90º

0º 180º

270º

360º

129


Reduction Formulas (The “Forehead Rule”) In example 17.1 we calculated sine, cosine and tangent for quadrant angles 0 , 90 ,

180 , 270 , and 360 . Here we describe the way how to simplify sine and cosine of angles when we add (or subtract) angle θ to (from) quadrant angles. In other words we will simplify the following expressions:

( ) ( ) ( ) ( )sin 90 , cos 180 , sin 270 , sin 360θ θ θ θ± ± ± ± .

Formulas that simplify these expressions are called reduction formulas. For example, it is not difficult to get that ( )sin 90 cosθ θ− = and ( )cos 90 sinθ θ− = (sine of an angle and cosine of complement angle are equal). Another example is

( )cos 180 cosθ θ+ = − . It is possible to analyze each of such expressions separately, and get all reduction formulas (there is total eight of them). Instead, we suggest a simple rule to get such formulas. We call this rule the Forehead Rule. Forehead Rule works like this. To get the reduction formula assume that angle θ is acute. We need to answer two questions:

1) Should we put a minus sign on the right side of the formula? 2) Should we change the sine to cosine and/or vice versa?

To answer the first question, determine the quadrant in which angle under consideration lies. Based on the quadrant, determine the sign of trig function (as described above). To answer the second question, move your head along the axis on which the quadrant angle lies. In doing this you automatically get answer “yes” or “no”. Example 16.2. Get reduction formulas for

( ) ( ) ( )sin 90 , cos 180 , sin 270θ θ θ+ + + . Solution. For ( )sin 90 θ+ :

1) Angle 90 θ+ lies in 2nd quadrant. Here sine is positive, so minus sign is not needed.

2) Move your head along vertical axis (where 90 angle is located) and you get the answer “yes”, so change sine to cosine. Final answer: ( )sin 90 cosθ θ+ = .

For ( )cos 180 θ+ :

1) Angle 180 θ+ lies in 3rd quadrant. Here cosine is negative, so minus sign is needed.

2) Move your head along horizontal axis (where 180 angle is located) and you get the answer “no”, so do not change cosine to sine. Final answer: ( )cos 180 cosθ θ+ = − .

For ( )sin 270 θ+ :

1) Angle 270 θ+ lies in 4th quadrant. Here sine is negative, so minus sign is needed.

130


2) Move your head along vertical axis (where 270 angle is located) and you get the answer “yes”, so change sine to cosine. Final answer: ( )sin 270 cosθ θ+ = − .

Special cases of reduction formulas (when quadrant angle is 0 ) are

( )sin sinθ θ− = − (odd property of sine)

( )cos cosθ θ− = (even property of cosine) Reference Angle This is a useful tool to reduce calculation of trig functions of arbitrary angles to acute angles.

Definition. Let θ be an arbitrary angle in standard position. Angle rθ is called the reference angle to θ , if it satisfies three conditions:

1) Terminal side of rθ coincides with the terminal side of θ . 2) Initial side of rθ is horizontal (it coincides with either the positive or negative

parts of the x-axis). 3) Angle rθ is acute angle.

Let’s see how reference angle rθ looks like depending on the quadrant in which original angle θ is located.

1) Angle θ is located in the first quadrant. Then rθ coincides with θ : rθ θ= .

2) Angle θ is located in the second quadrant. Then 180rθ θ= − : 3) Angle θ is located in the third quadrant. Then 180rθ θ= − :

rθ

180

θ

rθ

180

θ

131


4) Angle θ is located in the fourth quadrant. Then 360rθ θ= − : A reference angle is useful because up to the sign, the values of any trig function of θ coincide with the values of the same trig function for the reference angle rθ , and rθ is always an acute angle. You can check this using the reduction formulas described above.

Main Property of Reference Angle

Hence, to calculate the value of a trig function, it is enough to find the sign of the function and calculate the value of trig function of the reference angle.

Example 16.3. Calculate cos120 .

Solution. Angle 120 is located in the 2nd quadrant, so cos120 0< . This is the case 2)

in the pictures above. Reference angle 180 120 60rθ = − = . We have 1cos 602

= .

Therefore,

1cos120 .2

= −

Example 16.4. Calculate sin 225 .

Solution. Angle 225 is located in the 3rd quadrant, so sin 225 0< . This is the case 3)

above. Reference angle 225 180 45rθ = − = . We have 2cos 45

2= . Therefore,

2sin 225 .2

= −

Example 16.5. Calculate tan 330 .

Solution. Angle 330 is located in the 4th quadrant, so tan 330 0< . This is the case 4)

rθ

360

θ

The absolute value of any trig function of any angle is equal to the value of the same trig function of the reference angle.

132


above. Reference angle 360 330 30rθ = − = . We have 3tan 30

3= . Therefore,

3tan 330 .3

= −

Example 16.6. Find the values of the other five trig functions, if 5cos6

θ = − and

0tan >θ .

Solution. For reference angle rθ , 5cos6rθ = . Let’s draw a right triangle, using definition

of cos rθ as ratio of adjacent side to hypotenuse:

By the Pythagorean theorem, vertical leg of this triangle is 2 26 5 11− = . From here, 11sin6rθ = and

11tan5rθ = . Since cos 0θ < and 0tan >θ , angle θ lies in the 3rd

quadrant. Therefore, 11sin6

θ = − and 11tan5

θ = . Other three trig functions are:

1 5 5 11 1 6 1 6 6 11cot , sec , csc

tan 11 cos 5 sin 1111 11θ θ θ

θ θ θ= = = = = = = − = − .

It is possible to define trig function using a circle with an arbitrary radius r (not only a unit circle with r = 1). Namely, sine, cosine and tangent of any angle θ (in a standard position), which has point A(a, b) on its terminal side are:

2 2sin , cos , tan , .b a b r a br r a

θ θ θ= = = = +

Note. In the above formulas, the radius r is the distance from the point A(a, b) to the origin.

Example 16.7. Find the value of the six trig functions of the angle θ if point (2, 3)− lies on the terminal side of angle θ , and θ is in standard position.

Solution. We have a = 2, b = – 3. Using the above formulas,

5

6

rθ

133


2 2 2 22 ( 3) 13,r a b= + = + − =

3 3 13 2 13 3 3sin , cos , tan13 13 2 213

b a br r a

θ θ θ− −= = = − = = = = = − .

Other three trig function are

1 13 1 13 1 2csc , sec , cot .sin 3 cos 2 tan 3

θ θ θθ θ θ

= = − = = = = −

134


Exercises 16 In exercises 16.1 and 16.2, get the reduction formulas to the given expressions. 16.1. a) ( )sin 180 θ+

b) ( )cos 90 θ−

16.2. a) ( )cos 180 θ−

b) ( )sin 270 θ− In exercises 16.3 and 16.4, find the reference angles to the given angles. 16.3.

a) 130

b) 320

c) 250

d) 85

e) 130−

16.4. a) 200

b) 10

c) 310

d) 100

e) 210− In exercises 16.5 and 16.6, reference angle of angle θ and quadrant in which angle θ is located are given. Find angle θ in the interval from 0 to 360 . 16.5.

a) 40 , quadrant III

b) 70 , quadrant II

c) 50 , quadrant IV

d) 20 , quadrant I

16.6. a) 40 , quadrant IV

b) 70 , quadrant I

c) 50 , quadrant II

d) 20 , quadrant III In exercises 16.7 and 16.8,

1. Determine the quadrant in which angle is located. 2. Find the reference angle. 3. Calculate the exact value without using a calculator.

16.7.

a) sin 210

b) cos300

c) tan135

16.8. a) sin 315 b) cos150

c) tan 240

135


In exercises 16.9 and 16.10, use given information to determine the quadrant in which the angle θ is located and find the values of the five remaining trig functions. 16.9.

a) 2sin3

θ = − and tan 0θ <

b) 2cos5

θ = − and sin 0θ >

c) 3tan5

θ = and cos 0θ <

16.10.

a) 4sin7

θ = and cos 0θ <

b) 5cos8

θ = − and tan 0θ >

c) 7tan4

θ = − and sin 0θ <

In exercises 16.11 and 16.12, coordinates of a point are given. Find the values of six trig functions of an angle in the standard position for which the terminal side passes through this point. 16.11 a) ( 1, 2)− − b) (4, 5)− c) ( 3, 7)−

16.12. a) ( 3, 5)− b) ( 5, 6)− c) ( 4, 7)− −

136

Session 17: Solving Oblique Triangles – Law of Sines

Session 17

Solving Oblique Triangles – Law of Sines Oblique triangles are triangles that are not necessary right triangles. We are going to “solve” them. It means to find its basic elements – sides and angles, given some of them. First of all, let’s see what elements must be given. Obvious, if only angles are given and no sides, this info is not enough to determine sides since triangles with the same angles are similar and may have different sizes. So, at least one side must be given. We consider all possible cases when one, two or all three sides are given as well as some number of angles. More precisely, the following four cases are possible in solving triangles:

1) One side and two angles are given. 2) Two sides and an angle opposite to one of them are given. 3) Two sides and angle between them are given. 4) Three sides are given.

Main tools to solve these problems are two important theorems: Law of Sines and Law of Cosines. Here we consider Law of Sines and the first two problems. Law of Sines

In any triangle, the bigger side, the bigger opposite angle. However, sides are not proportional to the opposite angles. For example, in the right triangle 30 60− , if side opposite to 30 is a, then side opposite to 60 is 3a , which is not 2a. Law of Sines says that in any triangle sides are proportional to the sines of opposite angles. In other words, the ratio of any side to the sine of the opposite angle remains the same for all three sides in a given triangle. More formally, the following theorem is true. Theorem (Law of Sines). Consider triangle ABC: Then

C

cB

bA

asinsinsin

==

A

B

C

a

b

c

137


Proof. For simplicity, we consider only acute triangle (proof for obtuse triangle is slightly different, but similar). Let’s draw height h to the side b:

Height h breaks triangle ABC into two right triangles: ABD and BCD. Let’s consider sines of angles A and C:

In triangle ABD, chA =sin . Solve for h: Ach sin= .

In triangle BCD, ahC =sin . Solve for h: Cah sin= .

Equate the above two expressions for h: CaAc sinsin = . Divide both sides of this equation by sin sinA C⋅ and get

C

cA

asinsin

= . Similar ratio is true for the side b and angle B. The proof is completed.

Law of Sines works perfectly good for solving triangles for the case 1) above when a side and two angles of a triangle are given. In this case triangle is defined uniquely (we assume that the sum of given angles is less than 180 ). With no problem we can find the third angle by subtracting two given angles from 180 , and then use Law of Sines to find two other sides.

Example 17.1. Solve a triangle, if a = 14, 40=B , and 75=C .

Solution. We need to find angle A, and sides b and c.

1) 180 180 40 75 65A B C= − − = − − = .

2) By Law of Sines, B

bA

asinsin

= . From here, using calculator, we get

sin 14 sin 40 9.9sin sin 65

a BbA

⋅= = =

.

3) Again by Law of Sines, C

cA

asinsin

= . From here

sin 14 sin 75 14.9sin sin 65

a CcA

⋅= = =

.

A

B

C

a

b

c

D

h

138


Final answer: 65=A , b = 9.9, c = 14.9.

Using Law of Sines – Ambiguous Case

We consider how to solve a triangle for the case 2) above when two sides and an angle opposite to one of them are given. In this case a triangle is not always defined uniquely and we may face some difficulties to solve it. This is the ambiguous case. We will assume that the following data are given: sides a and b, and angle A opposite to side a.

Case: angle A is obtuse This is a simple case since only two options are possible: triangle does not exist or triangle is unique. To understand why, let’s draw angle A and mark side b on its slant side:

To get a triangle, we need to draw side a from the top point to meet with the horizontal side of angle A. Obvious, if side a is too short, it will not meet the horizontal side, and triangle does not exist:

For triangle to exist, side a must be greater than b. Then triangle is defined uniquely. We come up to the following

Proposition 17.1. Let two sides a and b, and obtuse angle A opposite to side a are given. Then

1) If a b≤ , triangle does not exist.

2) If a b> , triangle exists and unique.

Note. Conclusion in part 1) is also clear by the following reason: if a b≤ , then A B≤ . Angle A is obtuse, so B also must be obtuse. But triangle cannot have two obtuse angles.

Example 17.2. Solve a triangle, if a = 18, b = 14, and 130A = .

Solution. We need to find angles B and C, and side c. Using Law of Sines, we have

Bb

Aa

sinsin= . From here

A b

A b

a

139


sin 14 sin130sin 0.59618

b ABa

⋅= = =

.

Notice, that at this point we calculated sine of angle B, but not angle B itself. To restore the angle B from its sine, we can use the button 1sin− on a calculator similar to what we did in session 15 for right triangles. This button corresponds to the inverse sine. We have

1sin (0.596) 37B −= = .

Now it is easy to find angle C: 180 180 130 37 13C A B= − − = − − = . To find side c, we can use Law of Sines again:

sin 18sin13 5.3

sin sin sin sin130a c a Cc

A C A= ⇒ = = =

.

Final answer: 37 , 13 , 5.3B C c= = = .

Case: angle A is acute Similar to obtuse angle, let’s draw angle A and mark side b on its slant side:

To create a triangle, we draw side a from the top point. Four cases are possible here: 1) Side a is too short to meet with the horizontal side: Triangle does not exist.

2) Side a touches horizontal side exactly in one point:

We have right triangle which is unique.

3) Side a intersects horizontal side in two points:

A b

A b a

A b

a a

A b a

140


We have two triangles with sides a, b and angle A.

4) Side a is long enough, and to create a triangle, side a intersects horizontal side only in one point:

The triangle is unique. The top angle may be acute or obtuse. How can we distinguish the above four cases using the values of sides a, b and angle A? Take a look at this picture In your mind, draw side a from the top point. You can see that if a < h, side a is too short and triangle does not exist. If a = h, we can draw only one right triangle. If a is between h and b: h < a < b, we can draw side a on both sides (left and right) of the height h, and we have two triangles. Finally, if a b≥ , we can draw only one triangle. Notice that

sinh Ab= , so sinh b A= .

We come up to the following

Proposition 17.2. Let two sides a and b, and acute angle A opposite to side a are given.

1) If a b≥ , then triangle is unique. This triangle may be acute or obtuse.

2) If a b< , denote sinh b A= . Three cases are possible:

a) If a < h, then triangle does not exist. b) If a = h, then triangle is unique. It is a right triangle. c) If a > h, then there are two triangles. Both of them may be obtuse, or one is acute

and the other is obtuse (see Exercise 17.20). A practical way to use Proposition 17.2 is to directly apply the Law of Sines

Bb

Aa

sinsin= and solve this equation for sin B :

sinsin b ABa

= . Three cases are

possible when calculating sin B by the above formula:

1) sin 1B > . Because sin B cannot be greater than 1, triangle does not exist.

2) sin 1B = . Then 1sin (1) 90B −= = .The triangle is unique. It is a right triangle.

A b a

A

b b h

141


3) sin 1B < . Let sin B s= , then 1sin ( )B s−= . Angle B (as the inverse sine of a positive value) is always positive and acute. So, one triangle already exists. To understand whether another triangle exists, note that there is one more angle B′ such that sin sinB B′ = . This angle is supplemental to angle B: 180B B′ = − . Angle B′ is obtuse. Should we accept it as a second solution or reject it? Just compare a and b:

a) If a b≥ , then by Proposition 18.2, the triangle is unique, so the second triangle does not exist.

b) If a b< , then the second triangle exists having the obtuse angle 180B B′ = − .

Note. Another way to see whether second triangle exists, is to calculate supplemental angle 180B B′ = − in any case (regardless on which side is bigger: a or b). Then, if

180B A′ + < , accept B′ , and if 180B A′ + ≥ , reject it. Example 17.3. Let b = 20 and 30A = . Determine the number of triangles that satisfy the given conditions. If triangle exists, solve it.

1) a = 5 2) a = 10 3) a = 16 4) a = 25

Solution. Using the Law of Sines B

bA

asinsin

= , we have sinsin b ABa

= . From a

calculator (or just notice that 30 is a special angle), sin sin 30 0.5A = = , and

expression for sin B becomes 20 0.5sin B

a⋅

= , so 10sin Ba

= .

1) If a = 5, then 10sin 25

B = = . Because sine cannot be greater than 1, triangle does not

exist.

2) If a = 10, then 10sin 110

B = = and 1sin (1) 90B −= = . This is a right triangle. To

solve it, calculate angle C and side c.

90 90 30 60C A= − = − = . Side c can be found by the Pythagorean Theorem (notice that b is hypotenuse, and a and c are legs):

2 2 2 220 10 300 10 3c b a= − = − = = .

Final answer: 90 , 60 , 10 3B C c= = = .

3) If a = 16, then 10sin 0.62516

B = = and 1sin (0.625) 39B −= = . Another angle B′ ,

142


such that sin sinB B′ = is an obtuse angle supplemental to angle B: 180 180 39 141B B′ = − = − = . We accept it because

141 30 171 180B A′ + = + = < .

Another reason to accept B′ is that b > a. So, we have two triangles. Let’s solve them. It remains to find angle C and side c.

a) Triangle with angle 39B = . We have 180 180 30 39 111C A B= − − = − − = . By the Law of Sines,

sin 16sin111 20.87

sin sin sin sin 30a c a Cc

A C A= ⇒ = = =

.

b) Triangle with angle 141B = (we use letter B instead of B′ ).We have

180 180 30 141 9C A B= − − = − − = . By the Law of Sines,

sin 16sin 9 5.01


A C A= ⇒ = = =

.

Final answer: There are two triangles:

39 , 111 , 20.87B C c= = = .

141 , 9 , 5.01B C c= = = .

4) If a = 25, then 10sin 0.425

B = = and 1sin (0.4) 24B −= = . Another angle B′ , such

that sin sinB B′ = is supplemental to B and is obtuse angle:

180 180 24 156B B′ = − = − = . We reject it because

156 30 186 180B A′ + = + = > .

Another reason to reject B′ is that b < a and angle B′ cannot be obtuse (it should be less that angle A). So, we have only one triangle with angle 24B = . To solve the triangle, it remains to find angle C and side c.

180 180 30 24 126C A B= − − = − − = . By the Law of Sines,

sin 25sin126 40.45


A C A= ⇒ = = =

.

Final answer: 24 , 126 , 40.45B C c= = = .

143


Exercises 17

Round answers (where applicable) to the nearest tenth. For a triangle ABC, use the following notation: A, B, and C are angles, and a, b, and c are sides opposite to the corresponding angles. Similar notations are used for a triangle PQR with angles P, Q and R, and sides p, q and r, as well as for a triangle KLM with angles K, L and M, and sides k, l and m. In problems 17.1 and 17. 2, solve a trialgle PQR using the given information.

17.1. 47, 50 ,r P= = and 110Q =

. 17.2. 23, 28 ,p Q= = and 67R =

.

In problems 17.3 and 17. 4, solve a trialgle KLM using the given information.

17.3. 10, 15,k m= = and 40M =

. 17.4. 35, 20,l m= = and 70L =

.

17.5. Three friends, Alice, Bob and Carol, are camping in their own tents on a flat meadow in the woodland. Alice and Carol are 25 m apart. The angle going from Alice to Bob and Carol is 20 , and angle going from Bob to Alice and Carol is 110 . How far apart are Bob and Carol?

17.6. Back to previous problem. On the next trip, the three friends set up tents such that Alice and Bob are 15 m apart. The angle going from Bob to Alice and Carol is 70 , and angle going from Carol to Alice and Bob is 40 . How far apart are Alice and Carol?

17.7. On the next trip, the three friends set up tents such that Alice and Carol are 12 m apart. The angle going from Alice to Bob and Carol is 50 , and angle going from Carol to Alice and Bob is 110 . How far apart are Alice and Bob, and Bob and Carol?

17.8. A post is standing on the ground and is supported by two wires (one on each side going in opposite direction). The ends of the wires on the ground are 7 ft apart. The angle of elevation of one of the wires is 80 , and of the other wire is 65 . Find the length of each wire.

17.9. Back to problem 17.5. On the next trip, the three friends set up tents such that Alic and Bob are 23 m apart, and Bob and Carol are 18 m apart. The angle going from Carol to Alice and Bob 55 . What is the angle going from Alice to Bob and Carol?

17.10. Back to problem 17.5. On the next trip, the three friends set up tents such that Alic and Carol are 15 m apart, and Bob and Carol are 24 m apart. The angle going from Alice to Bob and Carol is 80 . What is the angle going from Bob to Alice and Carol?

144


17.11. On the next trip, the three friends set up tents such that Alic and Bob are 17 m apart, and Bob and Carol are 25 m apart. The angle going from Alice to Bob and Carol is 50 . What is the angle going from Bob to Alice and Carol?

17.12. On the next trip, the three friends set up tents such that Alice and Bob are 18 m apart, and Alic and Carol are 13 m apart. The angle going from Carol to Alics and Bob is 40 . What is the angle going from Alice to Bob and Carol?

17.13. On the next trip, the three friends set up tents such that Alice and Carol are 12 m apart, and Alice and Bob are 19 m apart. The angle going from Carol to Alice and Bob is 110 . How far apart are Bob and Carol?

17.14. On another trip, the three friends set up tents such that Alice and Carol are 19 m apart, and Bob and Carol are 25 m apart. The angle going from Alice to Bob and Carol is 50 . How far apart are Alice and Bob?

17.15. Nina intends to purchase a parcel of land in the shape of a triangle (say, ABC). She hired an assessor to measure the parcel. Nina told the assessor that it would be enough to provide her with minimum information such that she could calculate the remaining measurements (angles and sides) herself. The assessor submitted the following report: a = 370 ft, 91A = , b = 400 ft,. Recently Nina took a course of Trigonometry, and she decided to fire this assessor. Why?

17.16. Nina hired another assessor. This assessor submitted the following report: a = 370 ft, 71 , 115A B= = . Nina decided to fire this assessor as well. Why?

17.17. Nina then hired a third assessor. This assessor submitted the following report: a = 370 ft, 71 , 65 , 40A B C= = = . Nina again decided to fire this assessor. Why?

17.18. Nina hired another assessor. This assessor submitted the following report: a = 370 ft, 71A = , b = 500 ft, 42C = . Nina again decided to fire this assessor. Why?

17.19. Nina hired yet another assessor. This assessor submitted the following report: a = 370 ft, b = 500 ft, c = 120 ft. Nina decided to fire this assessor too. Why?

145


Challenge Problems

17.20. Nina hired another assessor. This assessor submitted the following report: a = 370 ft, 71A = , b = 400 ft. Nina decided to fire this assessor. Why?

17.21. Nina hired another assessor. This assessor submitted the following report: a = 370 ft, 71A = , b = 350 ft, 54C = . Nina decided to fire this assessor. Why?

17.22. Nina hired another assessor. This assessor submitted the following report: a = 370 ft, 61A = , b = 400 ft. This time Nina decided not to fire the assessor but to request additional information to be able to calculate remaining angles and sides. What additional info (non-numeric) is needed?

17.23. Two radar stations are located 15 miles apart. They detect an aircraft between them. The angle of elevation measured by the first station is 20 , and the angle of elevation measured by the second station is 40 . Find the altitude of the aircraft.

17.24. Back to previous problem, solve it in general form. Let d be the distance between radar stations, and let A and B be the angles of elevations to the aircraft from the stations. As before, the aircraft is between radar stations. Find the formula that expresses altitude of the aircraft in terms of A, B and d. In this case, it is possible to input this formula into a computer program to calculate the altitude instantly and track it during the aircraft flight between radar stations.

17.25. Back to problem 17.23. This time radar stations detect an aircraft on the right side of both stations. The distance between stations is 15 miles, and the angles of elevations to the aircraft are 20 and 70 . Find the altitude of the aircraft.

17.26. Back to previous problem, solve it in general form. Let d be the distance between the radar stations, and let A and B be the angles of elevations to the aircraft. As before, the aircraft is on the same side (left or right) of both radar stations. Find the formula that expresses altitude of the aircraft in terms of A, B and d.

17.27. A 2.5 m flagpole is NOT standing up straight on the ground. It is supported by two wires (one on each side going in opposite direction), each 3 m long. Both wires make a 55 angle with the ground. How far apart is each wire from the flagpole?

17.28. Let two sides a and b, and angle A opposite to side a in a triangle ABC are given. Prove the following statements:

a) If 0 tanb A a b< < < , then there are two obtuse triangles. b) If sin min( , tan )b A a b b A< < , then there are two triangles: one is acute

and the other is obtuse.

Hint: Use the picture located above Proposition 17.2, and consider two cases for angle A: 45A < and 45A > .

146

Session 18: Solving Oblique Triangles – Law of Cosines

Session 18

Solving Oblique Triangles – Law of Cosines In previous session, using the Law of Sines, we considered two problems on solving triangles from the total of four: when one side and two angles are given, and when two sides and an angle opposite to one of them are given. Here we consider the remaining two problems:

1) Two sides and the angle between them are given. 2) Three sides are given.

For both problems, a triangle is unique and we do not have an ambiguous case. Method to solve these problems is based on another important law in trigonometry: the Law of Cosines.

Note. Formally speaking, in problem 2) triangle does not exist, if one of the sides is greater or equal to the sum of two other sides. We will assume that this case will not happen.

Law of Cosines This law can be treated as generalization of the Pythagorean Theorem from right triangles to oblique ones. Consider the triangle

If angle C is not a right angle, we cannot conclude that 2 2 2c a b= + , so the Pythagorean Theorem is not true here. Instead, the following result is valid. Theorem (Law of Cosines). For any triangle,

Cabbac cos2222 −+=

Note. Consider the special case when 90=C (case of a right triangle). Then 090coscos == C and the above formula becomes 222 bac += which is exactly the

Pythagorean Theorem. Therefore, the Law of Cosines can be considered as a generalization of the Pythagorean Theorem to oblique triangles. Proof of the Law of Cosines. Similar to the proof of the Law of Sines, we consider only case of acute triangles. Let’s draw the height h to the side b:

A

B

C

a

b

c

147


Height h breaks the triangle ABC into two right triangles: ABD and BCD. Let’s write down the Pythagorean Theorem for each of them:

For the triangle ABD: 222 hADc += .

For the triangle BCD: 222 hDCa += .

Now subtract the second equation from the first one to eliminate 2h :

( )( )2 2 2 2c a AD DC AD DC AD DC− = − = + − .

Notice that AD + DC = b. From here AD = b – DC and

AD – DC = (b – DC) – DC = b – 2DC.

Formula for 2 2c a− becomes 2 2 2( 2 ) 2c a b b DC b bDC− = − = −

or 2 2 2 2c a b bDC= + − .

Now write down the definition of Ccos from the triangle BCD: a

DCC =cos . From

here CaDC cos= . Substitute this expression into the above formula for 2c :

Cabbac cos2222 −+= . The theorem is proved. Note. In this theorem, we have expressed side c through sides a, b and the angle C that is between them. Since all three sides play the same role, no one has any privileges against the others. Therefore, we can write similar expressions for the sides a and b:

Abccba cos2222 −+= and Baccab cos2222 −+= .

The Law of Cosines allows us to express cosine of any angle through three sides. To do this, just solve the above equations for cosines:

bcacbA

2cos

222 −+= ,

acbcaB

2cos

222 −+= ,

abcbaC

2cos

222 −+= .

As we mentioned, using the Law of Cosines we can solve triangles for the cases 1) and 2) indicated above. We will also use the property 180=++ CBA .

A

B

C

a

b

c

D

h

148


Case 1. Two sides and the angle between them are given.

Example 18.1. Solve a triangle, if a = 50, b = 15, and 55C = . Round the answers to the nearest tenth.

Solution. We need to find side c, and angles A and B. 1) By the Law of Cosines

2 2 2 2 22 cos 50 15 2 50 15 cos55c a b ab C= + − = + − ⋅ ⋅ ⋅ .

Using a calculator, cos55 0.5736= and

2 2500 225 1500 0.5736 1864.6c = + − ⋅ = .

1864.6 43.2c = = .

2) .

Using a calculator, ( )1cos 0.3167 108.5A −= − = .

3) 180 180 108.5 55 16.5B A C= − − = − − = .

Final answer: c = 43.2, 108.5A = , 16.5B = . Note. In solving problems for Case 1, it is possible in step 2) to use the Law of Sines instead of the Law of Cosines. However, you need to be very careful when using button

1sin− on calculator. This button always gives only an acute angle, but the actual angle may be obtuse. To avoid possible mistake, we recommend, when using the Law of Sines to calculate the angle, do not start with the angle opposite to the biggest side, because this angle may be obtuse. Always start from another angle that is definitely acute.

See, what may happen if you do not follow this advice. Let’s return to Example 18.1, and try to use the Law of Sines in step 2) to find angle A, which is opposite to the largest side a = 50:

We have sin sin

a cA C= . From here

sin 50sin 55 50 0.8192sin 0.9481

43.2 43.2a CA

c⋅

= = = =

,

and, using the calculator, ( )1sin 0.9481 71.5− = .

So, it looks like 71.5A = . However, this answer is wrong. You can check it by calculating the angle 180 180 71.5 55 53.5B A C= − − = − − = and using the Law of Sines:

50 52.7

sin sin 71.5a

A= =

, but 15 18.7

sin sin 53.5b

B= =

.

The correct answer is the supplemental obtuse angle 108.5 180 71.5= − .

149


When using the Law of Cosines, you do not always start with 2c . You need to start from the side for which the opposite angle is given. The following example demonstrates it. Example 18.2. Solve a triangle, if b = 12, c = 15, and 25A = . Round the answers to the nearest tenth.

Solution. We need to find a, B and C. 1) Because angle A is given, we start with its opposite side a:

2 2 2 2 22 cos 12 15 2 12 15 cos 25a b c bc A= + − = + − ⋅ ⋅ ⋅ .

Using the calculator, cos 25 0.9063= and 2 144 225 360 0.9063 42.73a = + − ⋅ = .

.

2) Let’s find angle B using the Law of Sines. This is safe because the opposite side b is not the largest one (see Note above). We have

( )1sin 12sin 25sin 0.775 sin 0.775 50.8sin sin 6.54

a b b AB BA B a

−= ⇒ = = = ⇒ = =

.

3) 180 180 25 50.8 104.2C A B= − − = − − = .

Final answer: 6.5, 50.8 , 104.2a B C= = = . Case 2. Three sides are given.

We only need to find three angles. Using the Law of Cosines, we can start with any side. We recommend to start with the biggest side and find the opposite angle. In doing this, we guarantee that the other two angles are acute, and to find them we can use either the Law of Cosines again or the Law of Sines (without making mistake indicated in the Note above). Here are our general recommendations:

1) When using the Law of Sines, start with the smallest side. 2) When using the Law of Cosines, start with the biggest side.

Example 18.3. Solve a triangle, if a = 12, b = 20, c = 17. Round the answers to the nearest tenth.

Solution. We need to find angles A, B and C. 1) According to the above recommendation, we use the Law of Cosines starting with the largest side b = 20.

Baccab cos2222 −+= . To find angle B, you can directly substitute given sides into this formula, or first solve it for cos B . We solve for cos B first

150


( )2 2 2 2 2 2

112 17 20cos 0.081, cos 0.081 85.42 2 12 17

a c bB Bac

−+ − + −= = = = =

⋅ ⋅ .

2) To find angle A, let’s use the Law of Sines

( )1sin 12sin85.4, sin 0.598, sin 0.598 36.7sin sin 20

a b a BA AA B b

−= = = = = =

.

3) 180 180 36.7 85 57.9C A B= − − = − − = .

Final answer: 36.7 , 85.4 , 57.9A B C= = = .

Example 18.4. Justify the following method to check whether a triangle with given sides a, b, and c is an acute, an obtuse or a right triangle:

Let c be the biggest side of the triangle. Calculate the value 222 cbaE −+= .

1) If E > 0, then the triangle is acute. 2) If E < 0, then the triangle is obtuse. 3) If E = 0, then the triangle is right.

Solution. Using the Law of Cosines, abE

abcbaC

22cos

222

=−+

= .

1) If E > 0, then 0cos >C and 90<C . Since c is a biggest side, C is a biggest angle. Therefore, two other angels are also less than 90 and the triangle is acute.

2) If E < 0, then 0cos <C and 90>C . The triangle is obtuse.

3) If E = 0, then 0cos =C and 90=C . The triangle is right.

Specific example of using this method was given in session 12, example 12.2.

Note. You can use the result of Example 18.4 (if the three sides of the triangle are given) to check what kind of triangle you have before calculating the angles.

151


Exercises 18

Round answers (where applicable) to the nearest tenth. For a triangle PQR, use the following notation: P, Q, and R are angles, and p, q, and r are sides opposite to the corresponding angles. Similar notations are used for a triangle KLM with angles K, L and M, and sides k, l and m.

In problems 18.1 and 18. 2, solve the trialgle PQR using the given information.

18.1. 20, 30,q r= = and 65P =

. 18.2. 35, 70,p r= = and 130Q =

.

In problems 18.3 and 18. 4, solve the trialgle KLM using the given information.

18.3. 10, 15, 20k l m= = = . 18.4. 12, 8, 5k l m= = = .

18.5. Eli and Ben came to the forest to pick some wild flowers. They started from the same point and each walked in a straight line at an angle of 40 relative to each other. Every minute they call out to each other to avoid being lost. A sound in this forest can be heard from up to 60 m away. After 10 minutes, Eli has walked 80 m and Ben has walked 70 m. Can they hear each other at that time?

18.6. Lillian wants to measure the distance between two trees that are on opposite sides of a small pond. She started at one of the trees and walked 240 ft in a straight line along the pond. Then she turned at 115 toward the second tree and walked another 310 ft until she reached the second tree. What is the distance between the trees?

18.7. Three friends, Alice, Bob and Carol, are camping in their own tents on a flat meadow in the woodland. Alice and Carol are 10 m apart, and Alice and Bob are 15 m apart. The angle going from Alice to Bob and Carol is 50 . What is the angle going from Bob to Alice and Carol?

18.8. Back to previous problem. On the next trip, the three friends set up tents such that Alice and Bob are 15 m apart, Bob and Carol are 20 m apart, and the angle going from Bob to Alice and Carol is 110 . Find the angle going from Carol to Alice and Bob.

18.9. A small airplane is 60 miles from the airport and is going down for landing with some angle of depression. However its navigation device malfunctions and incorrectly shows the distance to be 65 miles. The dispatcher noticed this mistake and figured out that if airplane continues on its current course, it will end up 16 miles from airport. By how many degrees should dispatcher adjust the airplane heading?

152


18.10. An aircraft is making a flight to airport A. At some point a pilot receives information that due to bad weather, airport A is closed, and he needs to fly to airport B. At that moment, the aircraft is 520 mi apart from airport A, and is 650 mi apart from airport B. The distance between airports A and B is 570 mi. By what angle should the pilot change the course of the aircraft to fly to airport B?

Challenge Problem

18.11. After hurricane Sandy, a small tree was leaning. To keep it from falling, it was nailed by a 7-foot strap into the ground 5 feet from the base of the tree. The strap was attached to the tree 4 feet along the tree from the ground. By what angle from the vertical position was the tree leaning?

153

Session 19: Radian Measure of Angles

Session 19

Radian Measure of Angles Most people familiar with the degree measure of angles. We already mentioned in session 15 that if we cut a round pizza pie (theoretically, of course) into 360 slices, the angle in one slice is of one degree (and this is a very tiny piece, so almost nothing to eat). But why the number 360 is used for the degree measure of angles? This number was introduced by astronomers in ancient Babylon (at least 3000 B.C.). No one knows for sure why they settled for this number. At those times, it was already known that the yearly cycle consists of 365 and 1/4 days, even though astronomers didn’t know yet that the earth revolves around the sun. It is reasonable to guess that they just rounded 365 and 1/4 to 360 because the number 360 has many more divisors. In other words, the number 360 can be divided into whole parts much better than 365. From this point of view, we could treat one degree angle as one day related to entire year. In any case, it’s clear that angle measure based on the number 360 is artificial. It looks similar to the decimal system which is also an artificial one since it was introduced only because we have 10 fingers on our hands. In math, and especially in computer science, it is used more convenient systems like binary or octal which have as bases powers of two. These systems could be considered as natural ones. And how about measurement of angles? Does some kind of natural measure of angles exist? The answer is “yes”. This measure is called the radian measure. To define the radian measure, consider an angle as a central angle in a circle. It means that we draw a circle and put the vertex of the angle in its center: Of course, we can draw infinite many such circles. One of them is a unit circle (its radius is equal to 1). Using it, the radian measure (denoted as θ ) of the central angle in unit circle is the length s of the corresponding arc (arc between two radii):

Central angle

s=θ

θ

1

1

154


For any other circle (with arbitrary radius), by the proportionality, the ratio of the arc to the radius equals to the above arc of the unit circle. We come up to the following definition for arbitrary circle. Definition of Radians. Consider an angle as a central angle: we draw a circle with the center in its vertex. Let the radius and the corresponding arc of the circle be r and s accordingly. Then the radian measure θ of the angle is defined as the ratio of s to r:

From here, s rθ= ⋅ . We may say that the radian measure of a central angle is the number of radii that can fit in the corresponding arc; hence the term “radian”. In particular, a central angle is of one radian measure, if the length of the corresponding arc is equal to the radius: s = r. We may also say that a one-radian angle is an angle in a “curvilinear” equilateral triangle (sector) in which two sides are radii, the third side is an arc, and all three sides are equal.

From this point of view, it is easy to estimate the value of one radian. As we know, in a “normal” equilateral triangle all three angles are of 60 . In “curvilinear” equilateral triangle, the central angle should be a bit less than 60 because the opposite side is an arc (a curve). Below in example 20.1 we will calculate that 1 radian 3.57≈ . As we see, it is much better to cut our pizza pie by radians. In this case at least 6 people (360/57.3 ≈ 6) will have something to eat. At the first glance the radian measure may look a bit more complicated than the degree measure. However it is more useful in some problems in mathematics and science.

rs

=θ θ r

155


To understand the benefit of radian measure, let’s re-write the above formula rs

=θ as

rs ⋅= θ . As you see, using the radian measure, the connection between arc, angle and radius is very simple. For any other measure of angles (for example, for degrees) this connection is more complicated and has the form rks ⋅⋅= θ , where k is some numerical coefficient (we will show in example 20.4 below that for the degree measure,

017.0≈k ). Radian measure is different from all others by the simplest value k = 1. The main idea of the radian measure is to relate linear (length of the arc) and angular measurements in the simplest possible way. That’s why many mathematical and technical calculations are simpler when using radians. The idea of measuring angles by the length of the arc is credited to Roger Cotes in the early 1700s, an English mathematician who worked closely with Isaac Newton. But the term radian was first introduced only in the late 1800s by James Thomson, Ireland.

Let’s set up connection between the radians and degrees. Consider the angle of 360 . This angle corresponds to a full rotation around a circle. If we consider it as a central angle, the corresponding arc s is the entire circumference. Recall the formula for the circumference of a circle: 2s rπ= ⋅ . Compare this formula with the above rs ⋅= θ . By equating both, we get 2r rθ π⋅ = ⋅ . From here, πθ 2= . We see that angle 360 corresponds to π2 radians. This connection allows to express any degree measure in radians and vice versa. In particular, 180 corresponds to π radians. For any angle, let’s denote its degree measure as θ , and the radian measure as rθ . It is easy to set up connection between θ and rθ , if we use the proportion: 180 relates to π as θ relates to rθ Let’s call this proportion the main proportion.

Using cross-multiplication, we get 180 rθ π θ⋅ = ⋅ . From here we can express θ through rθ and vice versa:

rθπθ ⋅=

180, θπθ ⋅=

180r .

Note. You do not need to memorize these formulas. Just remember that 180 corresponds to π radians: 180 radπ=

and then use the main proportion.

180

r

θπ θ

=

156


Example 19.1. Express the angle of 1 radian in degrees.

Solution. The main proportion takes the form

1801r

θπ

=

By cross-multiplication, 180=⋅πθ . From here, 180 180 57.3

3.14θ

π= ≈ ≈

.

So, 1 radian 57.3≈ .

Note. If angle in radians is given in terms of π , there is no need to use proportion to convert this angle into degrees: simply replace π with 180. In this way we can say

immediately that 2π

is 90 , 32π

is 270 , 2π is 360 and so on.

Example 19.2. Express the angle of 512π

radians in degrees.

Solution. Replace π with 180 and you are done: 5 5 180 75 .12 12π ⋅= =

Example 19.3. Express the angle of 1 in radians.

Solution. The main proportion takes the form

180 1

rπ θ=

By cross-multiplication, πθ =⋅ r180 . From here, 3.14 0.017.180 180rπθ = ≈ ≈

So, 1 0.017≈ radians.

Example 19.4. Express the arc length of a central angle through the radius of the circle and the degree measure of the angle.

Solution. Let s, r, and θ be the arc length, radius, and degree measure of the central angle accordingly. Also, denote by rθ the radian measure of the angle. As we mentioned

above, rs r ⋅= θ , and θπθ ⋅=180r . From here, rs ⋅⋅= θπ

180. Using calculation

0.017180π

≈ , we can write the approximate formula rs ⋅⋅≈ θ017.0 .

157


rs r ⋅= θ

Note. Let’s recall again that for radian measure, connection between arc length s, radius r, and central angle rθ is the simplest:

For any other measure this relation is more complicated, for example for degrees, rs ⋅⋅≈ θ017.0 .

Using the main proportion, we can calculate the radian measure of special angles 30 ,

45 and 60 , as well as of quadrant angles 0 , 90 180 270 360 . The following table summarizes the calculations. Degrees 0 30 45 60 90 180 270 360

Radians 0 6π

4π

3π

2π

π 2

3π π2

In conclusion, let’s mark quadrant angles in degrees and radians on the unit circle. Compare left and right figures.

90º

0º 180º –180º

270º –90º

360º 2π 0

–π

–π/2 3π/2

π

π/2

Angles in degrees Angles in radians

158


Exercises 19

In exercises 19.1 and 19.2, convert given angles from radians to degrees. Round the answers to the nearest tenth. 19.1. a) 1.3 b) – 0.6

19.2. a) 2.4 b) – 0.8

In exercises 19.3 and 19.4, convert given angles from radians to degrees.

19.3. a) 29π

b) 310π

−

19.4. a)

415π

b) 65π

−

In exercises 19.5 and 19.6, convert given angles from degrees to radians. Round the answers to the nearest hundreds. 19.5. a) 140

b) 85−

19.6. a) 78

b) 237− In exercises 19.7 and 19.8, convert given angles from degrees to radians. Write the answers in terms of π . Do not round the answers. 19.7. a) 120

b) 150−

19.8. a) 330

b) 225− In exercises 19.9 and 19.10, angle θ is given in radians. Without using a calculator, find the exact values of sine cosine and tangent of angle θ .

19.9. a) 43πθ =

b) 4πθ = −

c) 56πθ =

19.10. a)

6πθ = −

b) 23πθ =

c) 34πθ = −

In exercises 19.11 and 19.12, r is the radius of a circle, and θ is the central angle. Find the length of the arc bounded by angle θ . Round your answer to the nearest tenth. 19.11. a) r = 4.2 ft, 2.3θ = (rad)

b) r = 1.3 cm, 80θ =

19.12. a) r = 1.7 m, 0.3θ = (rad)

b) r = 2.5 in, 115θ =

159


Challenge Problems

19.13. The angles of a triangle are in the ratio of 3:4:5. Express these angles in radians.

19.14. The diameter of a Ferris wheel is 16 m. The spokes connecting two consecutive

cabs to the center of the wheel make an angle of 8π

. How many cabs are on the

wheel? What is the length of the arc between two consecutive cabs?

19.15. Nick is running around a circular track of radius 30-meters. Esther is standing at the center and observing him. She found that she turned 6 radians in one minute. What was the Nick’s speed?

19.16. A fly sat on the top of the second hand of a large clock and rode 48 cm until Lillian swatted it away. How long was the fly riding if the length of the second hand is 1.5 m?

160

Session 20: Graph and Basic Equation for Sine

Session 20

Graph and Basic Equation for Sine In this and the next session, we will consider three basic trig functions: sine, cosine and tangent. For these function, we construct graphs, and solve the three basic (or simplest) equations: sin , cosx a x a= = , and tan x a= . We call these equations basic because the solution of many more complicated equations can be reduced to them. In this section, we focus on the sine, and in the next, on the cosine and tangent. We will use the radian measure.

Function xy = sin( )

Let’s recall the definition of the sine for an arbitrary angle: we draw the angle in the standard position in the system of coordinates with the unit circle, and consider the point of interception of the terminal side of the angle with the unit circle. Sine is the second (vertical) coordinate of this point. To draw graph of sine, we will move along the unit circle, starting with the rightmost position, and observe how the vertical coordinates of the points on the unit circle change from quadrant to quadrant. Obviously, in the first quadrant, the vertical coordinate (i.e. sine) increases from zero to one:

To graph the sine, we will use a different system of coordinates in which we mark the angle on the horizontal axis (we will use the letter x instead of θ ), and mark sin x on the vertical y-axis. If you pick up several values of the angles x in the first quadrant (i.e. from 0 to / 2π ), calculate sin x , and plot the points in the system of coordinates, you will see that the sine does not increase along a strait line. Instead, it increases along the curve: Graph of sine in the first quadrant

y

x 1 0

•

θ sinθ = vertical coordinate

1

/ 2π x 0

1

161


Entire graph of the function xy sin=

In similar way, in the second quadrant (from / 2π to π ), sine decreases from 1 to 0: Graph of sine in the first and second quadrants Continue moving along the unit circle, we see that in the third quadrant (from π to 3 / 2π ) sine decreases from 0 to –1, and in the fourth quadrant (from 3 / 2π to 2π ) sine increases from –1 to 0. At this point, we get the graph of sine for one full cycle (we also say, on one period interval):

Graph of sine on one period interval [ ]0, 2π If we continue to move around the unit circle in either direction (positive or negative), we will expand the graph of sine to the entire number line, i.e. for all the values of x from – ∞ to + ∞: You can see that the domain of sine (possible values of x) is the interval (– ∞, + ∞) and the range (possible values of y) is [–1, 1]. Sine is periodical function with a period of 2π . It means that the sine repeats itself on each interval of the length 2π . More formally, for any x sin( 2 ) sin( )x xπ+ = (Periodic property of sine)

Also, the graph is symmetrical with respect to the origin. Algebraically, it means that

162


sin( ) sin( )x x− = − (Odd property of sine)

Solving Basic Equation x asin( ) = on One Period Interval [ )0, 2π

Notice that the right point 2π is not included in the above interval. The reason is that this point corresponds to the angle of 0, which is already taken for the left point of the interval. In this interval, equation sin( )x a= may have zero, one or two solutions depending on the value of a. More precisely, the following statements are true.

Proposition 21.1. Consider the equation sin( )x a= in the interval [ )0, 2π . Then

1) If 1a > , the equation does not have solutions.

2) If 1a = , the equation has one solution.

3) If 1a < , the equation has two solutions.

We can check all three statements using a geometric interpretation of the equation sin( )x a= : we consider intersections of the horizontal line y = a with the graph of sine. Then the x-coordinates of the points of intersections are the solutions. By drawing this horizontal line, we can see three different locations of it, depending on three different values of number a.

1) 1a > . This inequality is equivalent to 1a > or 1a < − . Horizontal line y = a is located above or below the graph of sine, so no points of intersection, and no solutions.

2) 1a = . This equality is equivalent to a = 1 or a = –1. In both cases line

y = a touches the graph of sine only in one point in the interval [ )0, 2π :

Line y = 1 touches the graph at the point ( )/ 2, 1π . Therefore, the equation sin( ) 1x = has only one solution / 2x π= .

Line y = –1 touches the graph at the point ( )3 / 2, 1π − . Therefore, the equation sin( ) 1x = − has only one solution 3 / 2.x π=

3) 1a < . This inequality is equivalent to 1 1a− < < . Line y = a is located between lines y = –1 and y = 1 and intersects the graph of sine exactly into two points in the interval [ )0, 2π . In particular, if a = 0, equation sin( ) 0x = has two solutions:

0x = and x π= .

To find the roots of the equation sin( )x a= when 0a ≠ and 1 1a− < < , we can use the reference angle. (For review of reference angles, see session 16).

163


Steps to solve the equation sin( ) , 0, 1 1x a a a= ≠ − < < .

1) Set up the equation for the reference angle rx : sin( )rx a= . In other words, ignore the sign of the number a (always take a plus sign). Solve this equation for the reference angle: ( )1sinrx a−= . To calculate rx , you may use the button

1sin− on a calculator. 2) Determine the quadrants in which the angle x is located based on the sign of the

number a: If a > 0, then angle x is located in the 1st and 2nd quadrants. If a < 0, then angle x is located in the 3rd and 4th quadrants.

3) Using quadrants and a reference angle, find two solutions 1x and 2x of the equation sin( )x a= .

If a > 0, then 1 rx x= and 2 rx xπ= − (solutions are in the 1st and 2nd quadrants):

If a < 0, then 1 rx xπ= + and 2 2 rx xπ= − (solutions are in the 3rd and 4th quadrants):

Example 20.1. Solve the equation 2sin( ) 4 5x + = in the interval [ )0, 2π .

Solution. We can reduce this equation to the basic one by solving for sin( )x : 12sin( ) 1 sin( )2

x x= ⇒ = .

1 rx x=

2 rx xπ= −

rx

rx

2 2 rx xπ= −

rx

1 rx xπ= +

164


Here a = 1/2 > 0. Therefore, the equation for reference angle rx is the same as for x: sin( ) 1/ 2rx = . From here 1sin (1/ 2)rx −= . We can find rx with a calculator or using the

special value 1/2: 1sin (1/ 2) 30 / 6rx π−= = = . The original equation has two roots. One of them, 1x , is located in the 1st quadrant and coincides with the reference angle:

1 / 6rx x π= = . The second root 2x is in the 2nd quadrant:

2 / 6 5 / 6rx xπ π π π= − = − = .

Final answer: 5,

6 6π π

.

Example 20.2. Solve the equation 2sin( ) 2x− = in the interval [ )0, 2π .

Solution. Solving for sin( )x , we get the basic equation sin( ) 2 / 2x = − . Here

2 / 2 0a = − < . The equation for reference angle rx is sin( ) 2 / 2rx a= = . From

here, using a calculator or the special value 2

2, we can find that

1 2sin 452 4rx π−

= = =

. The original equation has two roots. One of them, 1x , is

located in the 3rd quadrant: 15

4 4rx x π ππ π= + = + = . The second root 2x is in the 4th

quadrant: 272 2

4 4rx x π ππ π= − = − = .

Final answer: 5 7,4 4π π

.

Example 20.3. Solve the equation 5sin( ) 1 3x − = in the interval [ )0, 2π . Round the answer to the nearest hundredth.

Solution. Solving for sin( )x , we get the basic equation 4sin( )5

x = . Here 4 05

a = > . The

equation for reference angle rx is the same as for x: 4sin( )5rx = . The value

45

is not a

special value, and, to find rx , use a calculator (make sure the calculator is in radian

mode): 1 4sin 0.935rx − = =

Original equation has two roots. One of them 1x is located

in the 1st quadrant and coincides with the reference angle: 1 0.93rx x= = The second root

2x is in the 2nd quadrant: 2 0.93 2.21rx xπ π= − = − = .

Final answer: { }0.93, 2.21 .

165


Example 20.4. Solve the equation 6sin( ) 7 2x + = in the interval [ )0, 2π . Round the answer to the nearest hundredth.

Solution. Solving for sin( )x , we get the basic equation 5sin( )6

x = − . Here 5 06

a = − < .

The equation for reference angle rx is 5 5sin( )6 6rx = − = . The value

56

is not a special

value and, to find rx , we use a calculator in the radian mode: 1 5sin 0.996rx − = =

. The

original equation has two roots. One of them, 1x , is located in the 3rd quadrant:

1 0.99 4.13rx xπ π= + = + = . The second root 2x is in the 4th quadrant:

2 2 2 0.99 5.29rx xπ π= − = − = .

Final answer: { }4.13, 5.29 .

166


Exercises 20 In all exercises, solve the given equations in the interval [ )0, 2π . Use radian measure. In exercises 20.1 – 20.4, do not round the answers. Write the answers in terms of π . 20.1. a) 2 2 sin( ) 1 1x − = 20.2. a) 2 3 sin( ) 1 2x − =

b) 2 3 sin( ) 4 1x + = b) 2sin( ) 3 2x + =

20.3. 2 3 sin( ) 3 1x − = 20.4. 3 2 sin( ) 4 2x − =

In exercises 20.5 and 20.6, round the answers to the nearest hundredth.

20.5. a) 4sin( ) 1 2x − = 20.6. a) 5sin( ) 2 1x − =

b) 6sin( ) 5 3x + = b) 7sin( ) 6 2x + =

167

Session 21: Graphs and Basic Equations for Cosine and Tangent

Session 21

Graphs and Basic Equations for Cosine and Tangent

In the previous session, we examined the graph and the basic equation for the sine function. Here we will study these topics for cosine and tangent. Namely, we will construct graphs for cosine and tangent, and solve the basic equations cos x a= and tan x a= . As in the previous session, we will use the radian measure.

Function xy = cos( )

We can proceed similar to the sine function. We will do this in brief form hoping that the reader can restore details yourself. By definition, cosine is the first (horizontal) coordinate of a point on the unit circle that corresponds to given angle θ : Moving around the unit circle from quadrant to quadrant, we can construct the graph of cosine, observing how the horizontal coordinate changes. Similar to the case of the sine function, we use another system of coordinates in which we mark the angle on the horizontal axis (we will use the letter x instead of θ ), and mark cos x on the vertical y-axis. In the first quadrant when angle runs from 0 to / 2π , cosine decreases from 1 to 0: Graph of cosine in the first quadrant In the second quadrant cosine continue to decrease from 0 to –1, in third quadrant it increases from –1 to 0, and, finally, in fourth quadrant increases from 0 to 1. Here is the graph of cosine at one full cycle (on one period interval) from 0 to 2π :

y

x 1 0

•

θ

cosθ = horizontal coordinate

1

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Graph of cosine on one period interval [ ]0, 2π

If we extend graph to the entire x-axis, we get the complete graph of cosine:

Entire graph of the function cosy x=

As for sine, the domain of cosine is (– ∞, + ∞), range is [–1, 1], and cosine is periodical function with the same period 2π . Graph of cosine is symmetric with respect to y-axis:

cos( ) cos( )x x− = (Even property of cosine). Note. It is also possible to get the graph of cosine by shifting the graph of sine to the left by / 2π using the reduction formula cos( ) sin( / 2)x x π= + .

Solving Basic Equation x acos( ) = on One Period Interval [ )0, 2π

Number of solutions for this equation is exactly the same as for sine.

Proposition 21.1. Consider the equation cos( )x a= on the interval [ )0, 2π . Then

1) If 1a > , the equation does not have solutions.

2) If 1a = , the equation has one solution in the interval [ )0, 2π :

For equation cos( ) 1,x = the only solution is 0x = .

For equation cos( ) 1,x = − the only solution is .x π=

3) If 1a < , the equation has two roots. In particular, if a = 0, roots of the equation

cos( ) 0x = are 1 / 2x π= and 2 3 / 2x π= . If 0a ≠ , both roots can be found

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using the reference angle ( )1cosrx a−= in the same way as we did for the equation sin( )x a= . Also, the following formulas can be used that are true for any values of a, such that 1a < :

11 cos ( )x a−= and 1

2 2 cos ( )x aπ −= − .

Note. We described two methods of solving the equation cos( )x a= : use steps similar to those described for the equation sin( )x a= , and the above formulas.

Example 21.1. Solve the equation 2 3 cos( ) 1 2x − = in the interval [ )0, 2π .

Solution. Solving this equation for cos( )x , we have 3cos( )

2 3x = or

3cos( )2

x = .

Using steps similar to equation sin( )x a= , we set up the equation for the reference angle

rx : 3 3cos( )

2 2rx = = . From here 1 3cos 302 6rx π−

= = =

. Given equation has

two roots which are located in the 1st and 4th quadrants:

11

3cos 302 6

x π− = = =

, and 2112

6 6x π ππ= − = .

Example 21.2. Solve the equation 2cos( ) 4 3x + = in the interval [ )0, 2π .

Solution. Solving this equation for cos( )x , we have 1cos( )2

x = − . Using the second

method indicated in the Note above, we get two solutions:

11

1 2cos 1202 3

x π− = − = =

, and 22 423 3

x π ππ= − = .

Function xy = tan( )

On unit circle in the system of coordinates, we can interpret tangent like this. On the right side of unit circle, draw vertical line and extend terminal side of the angle to meet with that line. Then tangent is the vertical coordinate of the point of interception. Here are pictures of tangent when angle is located in each of the quadrants:

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Angle is in 1st quadrant Angle is in 2nd quadrant Angle is in 3rd quadrant Angle is in 4th quadrant We will draw the graph of tangent in the way similar to as we did for the sine and cosine. Moving along the unit circle in the first quadrant, note that the tangent increases from zero to infinity, and its graph in the 1st quadrant is as follows:

Graph of tangent in the first quadrant

Line / 2x π= becomes vertical asymptote. Continue moving in 2nd quadrant, we get the picture Graph of tangent in the first and second quadrants

1 θ

tanθ

1 θ

tanθ

tanθ

θ 1

tanθ θ

1

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Moving in 3rd and 4th quadrants, we get graph of tangent on interval [ )0, 2π :

Graph of tangent on [ ]0, 2π interval Continue to move around unit circle in both directions, we can draw a complete graph of the tangent:

Entire graph of the function tany x=

We see that the graph consists of an infinite number of branches, and it has infinite number of vertical asymptotes. The graph is symmetrical about the origin, so the tangent is an odd function: ( ) ( )tan tanx x− = − . It repeats itself on an π -length interval, so the

tangent has a period π : ( ) ( )tan tanx xπ+ = .

Solving Basic Equation x atan( ) = on Interval [ )0, 2π

Any horizontal line y = a intersects the graph of a tangent on [ )0, 2π interval always at

two points, so on this interval the equation ( )tan x a= always has exactly two roots for any a. The roots can be found in the same way as we described above for the equation sin( )x a= , using the reference angle rx which is a solution of the equation

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( )tan rx a= : ( )1tanrx a−= . To calculate the reference angle, we can use the button 1tan− on the calculator.

Proposition 21.2. For any a, the equation ( )tan x a= has two roots in the interval

[ )0, 2π . If 0a = , the roots are

1 20,x x π= = .

If 0a ≠ , then using the reference angle ( )1tanrx a−= , we have

1) If 0a > , the roots are in the 1st and the 3rd quadrants and are

1 2,r rx x x xπ= = + .

2) If 0a < , the roots are in the 2nd and the 4th quadrants and are

1 2, 2r rx x x xπ π= − = − .

Note. You do not need to memorize these formulas. Just draw a unit circle and mark the corresponding angles, as we did above for the equation sin( )x a= . Example 21.3. Solve the equation ( )3tan 2 3 3x − = in the interval [ )0, 2π .

Solution. Solving the equation for ( )tan x we get the basic equation ( )tan 3x = . The

reference angle rx is the solution of the equation ( )tan 3 3rx = = :

1tan ( 3) 60 / 3rx π−= = = .

Roots of the given equation are located in the 1st and 3rd quadrants and are

1 / 3rx x π= = ,

2 / 3 4 / 3rx xπ π π π= + = + = .

Example 21.4. Solve the equation 4 tan( ) 5 1x + = in the interval [ )0, 2π .

Solution. Solving the equation for ( )tan x we obtain the basic equation ( )tan 1x = − . The

reference angle rx is the solution of the equation ( )tan 1 1rx = − = :

( )1tan 1 45 / 4rx π−= = = .

Roots of the given equation are located in the 2nd and 4th quadrants and are

1 / 4 3 / 4rx xπ π π π= − = − = ,

2 2 2 / 4 7 / 4rx xπ π π π= − = − = .

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Exercises 21 In all exercises, solve the given equations in the interval [ )0, 2π . Use radian measure. In exercises 21.1 – 21.6, do not round the answers. Write the answers in terms of π . 21.1. a) 2 2 cos( ) 3 5x + = 21.2. a) 2cos( ) 5 6x + =

b) 2 3 cos( ) 5 2x + = b) 2 cos( ) 4 3x + =

21.3. a) 3 tan( ) 2 3x + = 21.4. a) 3 tan( ) 1 2x − =

b) 3 tan( ) 6 3x + = b) 3 tan( ) 6 5x + =

21.5. 3 2 cos( ) 7 2x − = 21.6. 2 3 cos( ) 5 1x − =

In exercises 21.7 – 21.10, round the answers to the nearest hundredth.

21.7. a) 5cos( ) 2 1x − = 21.8. a) 4cos( ) 1 2x − =

b) 7 cos( ) 6 2x + = b) 6cos( ) 5 3x + =

21.9. a) 3 tan( ) 1 5x − = 21.10. a) 2 tan( ) 3 5x − =

b) 4 tan( ) 7 5x + = b) 3 tan( ) 4 2x + =

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Session 22: Trigonometric Identities and None-Basic Equations

Session 22

Trigonometric Identities and None-Basic Equations Trigonometric Identities

Let’s compare the two statements:

sin cos 1x x+ = and 2 2sin cos 1x x+ = .

They look pretty much similar. However, they are completely different. The first one is the equation and the second one is the identity.

As we already know, equation is a statement that is true only for some specific values of variable, and the main problem for equation is to solve this, which means to find these specific values. Such values are called solutions or roots of the equation. For example, the values 0x = and / 2x π= are roots of the equation sin cos 1x x+ = . We can check this by substituting these values in the equation. We will have sin 0 cos 0 0 1 1+ = + = and ( ) ( )sin / 2 cos / 2 1 0 1π π+ = + = , so the equation becomes a true statement. If we pick, for example, x π= , it is not a root, because by substituting π for x, the equation does not become a true statement: ( ) ( )sin cos 0 1 1 1π π+ = − = − ≠ . Below in example 22.7 we show that roots 0 and / 2π are the only roots of the equation sin cos 1x x+ = in the interval [ )0, 2π .

On the contrary, the second statement 2 2sin cos 1x x+ = is true for any value of x, no exceptions. We can check this using the definitions of sine and cosine as the vertical and horizontal coordinates of points on a unit circle: These coordinates (i.e. sin x and cos x ) together with the radius (which is equal to 1) form a right triangle if a point ( )sin , cosx x lies in the 1st quadrant. By the Pythagorean

Theorem, 2 2sin cos 1x x+ = . We can check that this statement is true for all quadrants. Statements like this are called identities. Here is the exact definition.

Definition. Statement ( ) ( )f x g x= , where f and g are two functions, is called the identity, if this statement is true for all values of variable x from the common domain of functions f and g .

1 0 x . sin x

cos x

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The main problem for identity is to prove it, not to solve. In general, it is not possible to give exact recipe how to prove an identity. Common guideline is to try modify one or both sides f and g of the identity to get the same expression. Below we consider several examples. Let’s start with identities that we call basic. They can be used to prove more complicated identities. Four of them are simply expressions of tan , cot , secx x x and csc x through sin x and cos x : Another identity that we have already proven above is: We call this the main identity (it is also called the Pythagorean identity). It allows us to express 2sin x through 2cos x and vice versa:

2 2sin 1 cosx x= − and 2 2cos 1 sinx x= − . From the main identity, we can derive two more identities that connect tan x with sec x , and cot x with csc x : just divide both sides of the main identity by 2cos x and 2sin x :

2 2

2 2 2

sin cos 1cos cos cos

x xx x x+ = and

2 2

2 2 2

sin cos 1sin sin sin

x xx x x+ =

From here we get

There are a lot of trig identities that can be derived from the basics. Let’s consider some examples.

Example 22.1. Prove the identity ( )2 2tan 1 2 tan secx x x+ = + .

Solution. ( ) ( )2 2 2 2tan 1 tan 2 tan 1 tan 1 2 tan sec 2 tanx x x x x x x+ = + + = + + = + .

Example 22.2. Prove the identity 2 22 2

1 1 sec cscsin cos

x xx x+ = ⋅ .

Solution. We will modify the left side to get the right side:

2 2

2 22 2 2 2 2 2 2 2

1 1 cos sin 1 1 1 csc secsin cos sin cos sin cos sin cos

x x x xx x x x x x x x

++ = = = ⋅ = ⋅

⋅ ⋅.

sin cos 1 1tan , cot , sec , csccos sin cos sin

x xx x x xx x x x

= = = =

2 2sin cos 1x x+ =

2 2 2 2tan 1 sec , 1 cot cscx x x x+ = + =

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Example 22.3. Prove the identity 2 2 4 4sin cos sin cosx x x x− = − .

Solution. This time we modify the right side using the formula ( ) ( )bababa +⋅−=− 22 :

( ) ( )4 4 2 2 2 2 2 2sin cos sin cos sin cos sin cosx x x x x x x x− = − ⋅ + = − .

Example 22.4. Prove the identity ( )21 sin tan sec1 sin

x x xx

+= +

−.

Solution. Here both sides look rather complicated and we modify both of them. To modify the left side, we multiply the numerator and the denominator by 1 sin x+ . Then using the identity 2 21 sin cosx x− = in the denominator, we get

( )( )( )( )

( )2 2

2 2

22 2

2 2 2

2 2

1 sin 1 sin 1 sin1 sin 1 2sin sin1 sin 1 sin 1 sin 1 sin cos

1 2sin sin sin 1sec 2 tancos cos cos cos cossec 2 tan sec tan .

x x xx x xx x x x x

x x xx xx x x x xx x x x

+ + ++ + += = =

− − + −

= + + = + ⋅ +

= + ⋅ +

Now let’s modify the right side of the original identity:

( )2 2 2tan sec tan 2 tan sec secx x x x x x+ = + ⋅ +

We’ve got the same expression as for the left side. Identity is proved. None-Basic Trigonometric Equations In two previous sessions we solved basic trig equations sin , cosx a x a= = and tan x a= . Now we consider slightly more complicated equations that can be reduced to basic. To solve some of the equations, we will use basic identities. All equations we will solve in radians and in the interval [ )0, 2π .

Example 22.5. Solve the equation 28sin 14sin 15 0x x+ − = . Round the answer to the nearest hundredth. Solution. This equation can be treated as a quadratic equation with respect to xsin . Using the notation sin x u= , the equation becomes quadratic with respect to variable u:

28 14 15 0u u+ − = . It can be solved by the method described in session 2: 1) Construct the reduced equation by multiplying the last coefficient –15 by the leading

coefficient 8: 2 14 15 8 0u u+ − ⋅ = or 2 14 120 0u u− − = . 2) Solve the above reduced equation: ( )( )6 20 0 6u u u− + = ⇒ = and 20u = − . 3) Get the roots of the original equation by dividing 6 and –20 by the leading coefficient

8:

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6 3 0.758 4

u = = = and 20 5 2.58 2

u −= = − = − .

Replacing u with sin x , we get two basic trig equations: sin 0.75x = and sin 2.5x = − . We can solve them in the same way as we did in the previous sessions.

1) Equation sin 0.75x = has two solutions in the interval [ )0, 2π :

1sin (0.75) 0.85x −= = and 1sin (0.75) 2.29x π −= − = .

2) Equation sin 2.5x = − does not have solutions because xsin cannot be less than 1− .

Final answer: there are two solutions 0.85x = and 2.29x = .

Example 22.6. Solve the equation 28sin 2cos 5 0x x− − = .

Solution. We can rewrite this equation in terms of cos x using the identity 2 2sin 1 cosx x= − . Substituting this expression into the original equation, we obtain

( )2 28 1 cos 2cos 5 0 8 8cos 2cos 5 0x x x x− + − = ⇒ − − − = ⇒ 2 28cos 2cos 3 0 8cos 2cos 3 0x x x x− − + = ⇒ + − = .

Similar to example 22.5, we can treat this equation as quadratic with respect to cos x . Letting cosu x= , we have the equation 28 2 3 0u u+ − = . It can be solved in the same

way as in example 22.5. We omit the details and get 1 0.52

u = = and 3 0.754

u = − = − .

Replacing u with cos x , we get two basic trig equation cos 0.5x = and cos 0.75x = − . Let’s solve them. 1) Equation cos 0.5x = has two solutions (which are special angles)

( )1cos 0.5 603

x π−= = = and 52

3 3x π ππ= − = .

2) Equation cos 0.75x = − also has two solutions (which can be find using the calculator). We round the answer to the nearest hundredth:

( )1cos 0.75 2.42x −= − = and ( )12 cos 0.75 3.86x π −= − − = .

Final answer: there are four roots: 5, , 2.42,

3 3x x xπ π= = = and 3.86x = .

Note. The first two roots are exact solutions, while the last two are approximations to two decimal places.

Example 22.7. Solve the equation 2sin tan 3 tan 0x x x⋅ + = .

Solution. We can factor out tan x : ( )tan 2sin 3 0x x + = . Now equation can be split

into two: tan 0x = and 2sin 3 0x + = . The first one is basic, and the second one can

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be written as basic: sin 3 / 2x = − . Let’s solve them. Both of them have two solutions.

For tan 0x = , 0x = and x π= .

For sin 3 / 2x = − we can use steps described in Session 20: Consider the equation for the reference angle rx :

sin 3 / 2 3 / 2 60 / 3.r rx x π= − = ⇒ = =

Since sin 0x < , angle x is located in the 3rd and 4th quadrants, and we have two solutions:

/ 3 4 / 3rx xπ π π π= + = + = and 2 2 / 3 5 / 3rx xπ π π π= − = − = .

Final answer: there are four solutions: 0, , 4 / 3, 5 / 3x x x xπ π π= = = = . Example 22.8. Solve the equation sin cos 1x x+ = .

Solution. Let’s square both sides:

( )2 2 2 2sin cos 1 sin 2sin cos cos 1x x x x x x+ = ⇒ + ⋅ + = .

Using the main identity 2 2sin cos 1x x+ = , we can simplify the above equation: 1 2sin cos 1 2sin cos 0 sin cos 0x x x x x x+ ⋅ = ⇒ ⋅ = ⇒ ⋅ = .

This equation can be split into two basic equations: sin 0x = and cos 0x = . Equation sin 0x = has two solutions 0x = and x π= . Equation cos 0x = also has two

solutions 2

x π= and

322 2

x π ππ= − = . So, it looks like the original equation has four

solutions:

30, , ,

2 2x x x xπ ππ= = = = .

However, this is not true. We need to be very careful, and check these values with the original equation. The reason is that when we square both sides of an equation, we can get additional roots that are extraneous to the original equation. We already saw that in session 8, examples 8.6 and 8.8. Let’s check the above four values.

The values 0x = , 2

x π= and x π= we already checked at the beginning of this

session: 0 and 2π

are roots, but x π= is not. Let’s check 32

x π= :

3 3sin cos 1 0 1 12 2π π + = − + = − ≠

. So,

32

x π= is not the root, and we reject it.

Final answer: there are only two roots 0x = and 2

x π= .

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Exercises 22

In exercises 22.1 – 22.10, prove the given identities. 22.1. 2 2 2cos tan 1 cosx x x⋅ = − 22.2. 2 2 2sin cot 1 sinx x x⋅ = −

22.3. cot cos csc sinx x x x⋅ = − 22.4. tan sin sec cosx x x x⋅ = −

22.5. coscsc sintan

xx xx

− = 22.6. sinsec coscot

xx xx

− =

22.7. tan tan 2csc

1 cos 1 cosx x x

x x− =

− + 22.8.

cot cot 2sec1 sin 1 sin

x x xx x− =

− +

22.9. cot cot 2sec

1 csc 1 cscx x x

x x− =

+ − 22.10.

tan tan 2csc1 sec 1 sec

x x xx x− =

+ −

22.11. 4 4

22

cos sin 2 seccosx x x

x−

= − 22.12. 4 4

22

sin cos 2 cscsinx x x

x−

= −

In exercises 22.13 – 22.26, solve the given equations in the interval [ )0, 2π . Use radian measure. If it is possible to find exact solutions, write them in terms of π . If it is not possible, round the answers to the nearest thousandth.

22.13. 2cos ( ) cos( )x x= 22.14. 2sin ( ) sin( )x x=

22.15. 2sin ( ) sin( ) 0x x+ = 22.16. 2cos ( ) cos( ) 0x x+ =

22.17. 24cos ( ) 1 1x − = 22.18. 26sin ( ) 1 4x + =

22.19. 22sin ( ) 3sin( ) 2x x− = 22.20. 22cos ( ) 7cos( ) 3x x+ = −

22.21. 23sin ( ) cos( ) 1 0x x− − = 22.22. 25cos ( ) sin( ) 1 0x x+ − =

22.23. 3sin( ) tan( ) 3 sin( )x x x⋅ = 22.24. cos( ) tan( ) 3 cos( )x x x⋅ =

22.25. sin( ) cos( ) 1x x− = 22.26. cos( ) sin( ) 1x x− =

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Part III

Exponential and Logarithmic Functions

181

Session 23: Logarithms

Session 23

Logarithms Consider the following problem. Suppose we have three numbers x, y and z, that are connected by the equation xy z= . How to solve this equation for x and for y?

It is easy to solve for y: raise both sides of the equation to the power 1x

, and get

( )1 11

1xxx x x xxy y y y y z

⋅= = = = = . So,

1xy z= .

It is important to understand that even we obtained the formula 1xy z= , this formula, in

general, does not give us a direct way (a finite sequence of arithmetic operations) to get

the exact answer. Actually, the expression 1xz only provides a notation of a specific

operation on x and z, and the question of how to perform this operation is another story (which is beyond the scope of this textbook).

Similar situation occurs when we want to solve the equation xy z= for x. In other words, we want to express power x in terms of the base y and number z. Of course, for some specific values of y and z, it is easy to do.

Example 23.1. Solve the equation 2 8x = .

Solution. This equation can be solved directly. Indeed, we can represent the number 8 as an exponent with a base of 2: 328 = . Then the equation takes the form 32 2x = . From here we immediately conclude that x = 3.

However, in general, we cannot solve the equation xy z= for x so easily. Consider, for example, equation 2 6.x = Because 22 4= and 32 8= , we can just estimate that x should be somewhere between 2 and 3. But where? We cannot indicate the exact value. At the end of this session we will be able to get an approximation. We will see in Example 23.8 that up to three decimals, 2.585.x ≈

In general, we may think of the solution x of the equation xy z= as a result of some specific operation that we perform on y and z. In other words, we consider x as some function of two variables y and z. Since we have a function, we need a notation for it. You may invent your own notation. For example, using the abbreviation “sol” for solution, we can write x = sol(y, z). In mathematics, the following notation is used:

log yx z= . We read this as the “logarithm (or, in short, log) of the number z with the base

y”. So, the solution of the equation xy z= with respect to x is log yx z= .

In the definition below, we simply change letter y to b and z to c.

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Definition of Logarithm. Let b be a positive number not equal to 1, and c be any positive number. Then logbx c= is the solution of the equation xb c= . In other words, the logarithm is a power to which we raise the base b to get the number c. Note. You may be wondering why the base 1b ≠ . Well, let’s b = 1, so we consider the equation cx =1 . If we raise 1 to any power, the result is still 1, so we have c = 1, and any number x satisfies the equation 11 =x . Therefore, in this case, the solution 1log 1x = does not make sense, since it can be any number. Another restriction is b > 0. This is to avoid problems with complex (not real) numbers. For example, ( )1/ 21 1− = − is not a real number, so we exclude negative base b. Also, we put the restriction on number c: c > 0. This is because xc b= , and positive b raised to any power is positive, so for non-positive c logarithm does not exist. In practice, often number 10 is used as the base of logarithms. Such logarithms are called common ones. Usually, for simplicity we drop the base 10 in the notation of common logs. So, cc 10loglog = Working with logs, it is often convenient to convert them into exponential expressions. If we denote given logarithm by x (i.e. logb c x= ), we can re-write it (by definition) as

xb c= . To be more comfortable with this technique, you may memorize the following “Circular Rule” for conversion: in logb c x= , take base b, raise it to power x, and you get c:

log xb c x b c= ⇔ =

This rule says that two statements: logbx c= and xc b= are equivalent.

Logarithms were invented by Scottish mathematician John Napier in early 1600, and the notation log was introduced by German mathematician Gottfried Leibniz in 1675. In some cases, it is easier to operate with exponential expressions than with logarithms. We will see this in the following example.

Example 23.2. Calculate or simplify a) 8log2

b) 100log

c) 0001.0log

d) log 1b

e) logb b

f) log nb b

g) logb cb

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Solution.

a) The problem for computing 8log2 is, in fact, the same as example 23.1, but written in a different form. Indeed, if 8log2=x , then, by the circular rule, 82 =x . From example 23.1 we have x = 3, so 38log2 = .

b) Let 100log100log 10==x . By the circular rule, 21010010 ==x , so log100 2x = = .

c) Let 0001.0log=x . Then 4100001.010 −==x , so x = log 0.0001 4= − .

d) Let log 1bx = . Then 01xb b= = . Therefore, x = log 1 0b = .

e) Let logbx b= . Then 1xb b b= = . Therefore, x = log 1b b = .

f) Let log nbx b= . Then x nb b= . Therefore, x = log n

b b n= .

g) At the first glance, the expression logb cb looks rather complicated. However, if you look closely at this, you will realize that it is actually a very simple. Indeed, if we denote logb c x= , then xb c= , so, logb c xb b c= = .

Note. Try to memorize answers of problems 23.2, d) and e):

For any base b, log 1 0b = and log 1b b = .

Example 23.3. Prove that 1log logbb

c c= − .

Solution. Let’s use letters x and y for the above logs: 1log , logbb

x c y c= = . Then

1 1 ,x

x yx b c b c

b b− = = = =

. From here, x yb b− = . So, y x= − . Therefore,

1log logbb

c c= − .

Basic properties of logarithms

Multiplication Rule: log ( ) log logb b bx y x y⋅ = + .

In words: logarithm of product is equal to the sum of logarithms.

The proof of this statement can be done in a manner similar to examples 23.2 and 23.3. Denote each of three logs by letters: log ( ), log , and logb b bA x y B x C y= ⋅ = = . Next, use the circular rule to convert them into exponents: Ab x y= ⋅ , Bb x= , and

.Cb y= Now, multiply the second and third equations: , or .B C B Cb b x y b x y+⋅ = ⋅ = ⋅ Compare this equation with Ab x y= ⋅ . From here, A B Cb b += , hence A = B + C, or log ( ) log logb b bx y x y⋅ = + .

184


Note. Before the era of calculators, there was a widely used device, so-called logarithmic ruler, or slide ruler, that allows to multiply numbers based on the Multiplication Rule for logarithms. Schematically speaking, this device works like this. It contains two rulers that allow us to convert numbers into logs and vice versa. To multiply two numbers, the device converts them into logs and adds them up. According to the Multiplication Rule, this sum is the log of the product. Then the device converts this log of product back to the product of the given numbers. So, the device does (physically) summation, but mathematically we get multiplication.

Example 23.4. Solve the equation 3)3(log)1(log 22 =−+− xx .

Solution. Using the Multiplication Rule, we can combine both logs in one: 3)3)(1(log2 =−− xx . From here, 82)3)(1( 3 ==−− xx . This is a quadratic equation that

can be written in standard form 0542 =−− xx . We can solve it by factoring: (x – 5)(x + 1) = 0, and we get two solutions: x = 5 and x = – 1. Let’s check these solutions with the original equation. Let x = 5. Then 3122log4log)35(log)15(log 2222 =+=+=−+− ,

so everything is OK with this. Now, let x = – 1. We obtain the logs of negative numbers: )2(log2 − and )4(log2 − . Such logs do not make sense. Therefore, we must reject the

value x = – 1. Final answer: the equation has the only solution x = 5.

Notes. 1) Example 23.4 shows that we need to be very careful when we get the final answer: we

must check the answer with the original equation.

2) Keep in mind that log log log ( )a b bx y x y+ ≠ ⋅ if a b≠ . Multiplication Rule is applicable only to logs with the same base.

Quotient (or Division) Rule: log log logb b bx x yy

= −

.

In words: logarithm of quotient is equal to the difference of logarithms. The proof is similar to that given for multiplication rule. The proof of this statement as well as others statements given below, are left as exercises.

Note. The above two rules: the multiplication and division rules for logs, can be considered as the inverse rules for multiplication and division of exponential expression. If we multiply exponential expressions, we add their powers (powers are logs), if we add logs, we multiply their numbers. If we divide exponential expression, we subtract their powers, if we subtract logs, we divide their numbers. In the next session, we will discuss more about this “inverse connection”.

Example 23.5. Solve the equation 1)74log()8log( =−−+ xx .

Solution. Using the Division Rule, we can represent the left part as the logarithm of the

quotient, and the equation takes the form 1748log =−+

xx . From here 1010

748 1 ==−+

xx .

185


Solving this equation, we get x = 2. Let’s check this answer with the original equation:

1011log10log)724log()82log( =−=−=−×−+ . So, x = 2 is a solution.

Power Rule: log logn

b bx n x= ⋅ .

In words: the logarithm of an exponential expression is equal to its power times the logarithm of its base (the base of the exponential expression, not the base of log). Example 23.6. Calculate without using a calculator:

6 6 63 log 3 log 75 2 log 10⋅ − + ⋅

Solution. Here we can use all three rules listed above. We can modify the first and third terms like this:

36 6 63 log 3 log 3 log 27⋅ = =

26 6 62 log 10 log 10 log 100⋅ = =

From here

6 6 6 6 6 6

26 6 6 6

3 log 3 log 75 2 log 10 log 27 log 75 log 10027 100log log 36 log 6 2 log 6 2.

75

⋅ − + ⋅ = − +⋅

= = = = ⋅ =

Some scientific calculators allow us to calculate logarithms with only specific bases: base 10 (common logs), and base e (this is a special constant number which we will discuss later in session 25). Logs with the base e are denoted with the symbol ln and are called the natural logarithms. So ln logex x=

To calculate logs with other bases, we need a way to convert logs from one base to another. The following rule allows us to do this.

Change-of-Base Rule: loglog

logd

bd

xxb

= .

If we need to calculate the logarithm with the base b, but our ability is limited only by the base d, we can make a conversion from base b to d using the Change-of-Base Rule, and then perform the calculations. If we put x = d, we get a special case of the Change-of-Base Rule:

1loglogb

d

db

= .

To prove the Change-of-Base Rule, denote logby x= and convert it into exponential form . Now apply log with the base d to both sides and use the power rule:

186


loglog log log loglog

y dd d d d

d

xb x y b x yb

= ⇒ = ⇒ = .

Example 23.7. Assume that you have a calculator that allows you to calculate common logs (logs with the base of 10). Calculate 5log3 .

Solution. Using the Change-of-Base Rule and a calculator, we have

465.1477.0699.0

3log5log5log3 ≈≈= .

Logarithms are often useful in solving exponential equations in which the powers of exponential expressions are unknown. Example 23.8. Solve the equation 2 6x = that we discussed at the beginning of this session. Approximate the solution to the nearest thousandth.

Solution. We take the logarithm of both sides of this equation to the base 10 (common

log): log 2 log 6.x = Using the Power Rule, log 2 log 6,x ⋅ = and log 6log 2

x = . This is “exact

answer”. Using a calculator, we can get a numerical approximation:

log 6 0.7781 2.585log 2 0.3010

≈ ≈ . So, 2.585.x ≈

Example 23.9. Solve the equation 53 12 =−x . Approximate the solution to the nearest hundredth.

Solution. As in example 23.8, we take log of both sides: 5log3log 12 =−x . Using the Power Rule, we have 5log3log)12( =−x . From here,

23.11477.0699.0

211

3log5log

21,1

3log5log2,

3log5log12 ≈

+≈

+=+==− xxx .

So, 23.1≈x .

Note. The method of taking the logarithm of both sides of a given equation, which we used in examples 23.8 and 23.9, is very often used for equations containing exponents. Theoretically, we can take the logarithm with any base. We used common logs (logs with the base 10) to be able to use a calculator. We could also use natural logs (logs with the base e).

187


Exercises 23

In exercises 23.1 and 23.2, convert the given statements from logarithmic form to exponential form.

23.1. a) 2log 16 4= 23.2. a) 4log 64 3=

b) 21log 4

16 = −

b) 41log 364

= −

In exercises 23.3 and 23.4, convert the given statements from exponential form to logarithmic form (use bases of exponential expressions as the bases of logs).

23.3. a) 32 8= 23.4. a) 43 81=

b) 3 128

− = b) 4 1381

− =

In exercises 23.5 and 23.6, calculate without using a calculator.

23.5. a) 3log 81 23.6. a) 4log 16

b) log1000 b) log10000

c) log 0.01 c) log 0.001

d) 12

log 64 d) 13

log 27

e) 61log36

e) 41log

16

f) 54log 4 f) 4

6log 6

g) ( )43log 9 3 g) ( )5

7log 49 7

In exercises 23.7 – 23.10, solve the given equation.

23.7. a) 3 27x = 23.8. a) 5 625x =

b) 3 1 1464

x− = b) 2 3 1636

x− =

c) 2 31 125

5

x+ =

c) 3 21 81

3

x+ =

188


23.9. a) 3 3log ( 4) log ( 2) 3x x− + + = 23.10. a) 4 4log ( 4) log ( 2) 2x x+ + − =

b) 5 5log (7 3) log ( 3) 2x x− − − = b) 2 2log (15 3) log ( 3) 3x x+ − + = In exercises 23.11 and 23.12, calculate without using a calculator. 23.11. 8 8 8log 32 2 log 6 log 9− ⋅ + 23.12. 3 log 20 2 log3 log 72⋅ + ⋅ −

In exercises 23.13 and 23.14, let lnu x= and lnv y= Write given explressions in terms of u and v.

23.13. ( )3 2ln x y 23.14. ( )4 5ln x y In exercises 23.15 and 23.16, let logu x= and logv y= . Write given explressions in terms of u and v.

23.15. 7

6log xy

23.16. 3

2log yx

In exercises 23.17 and 23.18, assume that a calculator allows us to calculate only common logs (logs with the base 10) and natural logs (logs with the base e). Calculate the given expressions using both types of logs. Round your answers to the nearest thousands. Compare the results when using log and ln. 23.17. 7log 8 23.18. 8log 7 In exercises 23.19 and 23.20, solve the given equations. Round your answers to the nearest thousands. 23.19. a) 6 9x = 23.20. a) 5 8x =

b) 3 27 4x+ = b) 5 34 7x− =

c) 4 3 5xe + = c) 3 4 6xe − =

Challenge Problems

23.21. Prove that ( )11log logb

b

aa

=

.


log ( ) log ( )c cx a x b p q+ + + = + , where p qb a c c= − + .

Prove that the only solution of this equation is px c a= − .

189

Session 24: Exponential and Logarithmic Functions

Session 24

Exponential and Logarithmic Functions We already studied some functions: quadratic functions (parabolas) and trigonometric functions. In this session, we will study exponential expressions and logarithms from the point of view of functions. For these functions, we will denote by the letter a any positive number, not equal to 1, and call it the base of a function.

Exponential Functions

We can treat the expression xa as a function of x: if we pick any number x, the expression will produce the value xay = . We can also write xaxf =)( . The domain of this function (the set of possible values of x) is the set of all real numbers (since any number x can be taken as a power, so there are no exceptions), but the range (the set of possible values of y) is the set of only positive numbers (since the value of xa cannot be negative number or zero). We are interested in the behavior of this function. It means that we want to know what happens with the value y when x takes some specific values, when x increases to positive infinity, or decreases to negative infinity. One of the ways to study a function is to visualize it, in other words, construct its graph. One point on the graph is easy to observe: if x = 0 then 10 == ay . So, for any base a, the graph of the exponential function

xay = passes through the point (0, 1) which is located on the y-axis. It turns out that the shape of the graph of the function depends whether base a is greater or less than 1. (Case a = 1 is not interesting since for any x, and the graph of y = 1 is a horizontal line).

1. Case a > 1. In this case, the larger x, the larger y. We say that function xay = increases, and it increases very fast. For example, if we take a = 2, we can construct the following tables of values of function 2xy = for non-negative and negative values of x:

x 0 1 2 3 4 5

xy 2= 1 2 4 8 16 32

Non-negative value of x

x – 1 – 2 – 3 – 4 – 5 xy 2=

212 1 =−

412 2 =−

812 3 =−

1612 4 =−

3212 5 =−

Negative values of x

Based on these two tables we can draw the graph of the function xy 2= :

190


Notice that when x goes to positive infinity (moving to the right), y also goes to positive infinity (moving up), and when x goes to negative infinity (moving to the left), y approaches to zero (approaches to the x-axis and never touches it). We say that the x-axis is the horizontal asymptote of the function xay = .

2. Case 0 < a < 1. In this case, the larger x, the smaller y. We say that function xay =

decreases. Let’s take as an example 21

=a . We can construct the graph of the function

x

y

=

21 in a similar way as we did above for the function xy 2= by creating the table

of its values. However, we can get this graph almost immediately, if we notice that

xx

x−==

2

21

21 . So, we actually need to graph the function xy −= 2 .

Let’s consider in general the relationship between the graphs of the functions )(xf and )( xf − . Points (x, y) and (–x, y) are symmetrical to each other with respect to the y-axis.

Therefore, the graphs of )(xf and )( xf − are also symmetrical to each other with respect to the y-axis. This means that if we have already drawn the graph of )(xf , then to get the graph of )( xf − , we can just reflect the graph of )(xf with respect to the y-axis.

We can apply the above reasoning to the function xx

y −=

= 2

21 and reflect the graph of

xy 2= with respect to the y-axis. Here is the resulting picture

xy 2=

191


As you can see, when x goes to negative infinity, y increases to positive infinity, and when x goes to positive infinity, y approaches to zero (but doesn’t equal to it), so the

x-axis is still horizontal asymptote of the function x

y

=

21 .

Logarithmic Functions Similar to exponential expressions, we can treat the logarithm xalog (with fixed base a) as a function of x: xy alog= . Its domain (the set of possible values of x) is the set of all positive numbers, and range (the set of possible values of y) is the set of all real numbers.

Note. Notice that the domain of xalog is the range of xa , and the range of xalog is the domain of xa . This is not a coincidence: we will see shortly that this is connected with the concept of inverse functions.

Example 24.1. Find the domain of the following functions:

a) log (3 1)ay x= + b) log (5 2 )ay x= −

Solution. Since the domain of the function xy alog= is the set of all positive numbers, to find domains of given functions, we just need to solve the inequalities that result by making expressions in parentheses greater than zero:

a) 3 1 0 3 1 1/ 3x x x+ > ⇒ > − ⇒ > − . So, the domain is the interval ( )1/ 3,− ∞

b) 5 2 0 2 5 5 / 2x x x− > ⇒ − > − ⇒ < . So, the domail is the inteval ( ), 5 / 2−∞

All logarithmic functions (for all bases a) have the same value of zero at x = 1: 01log =a .

So, the graphs of all logs pass through the same point (1, 0) on the x-axis. Similar to exponential functions, the shape of the graph of the function xy alog= depends whether a is greater or less than 1.

x

y

=

21

192


1. Case a > 1. In this case, as for an exponential function, the log function increases, but this time increases very slowly. Let’s take as an example a = 2. Here are tables of the values of the function xy 2log= for the values of x greater and less than 1.

x 1 2 4 8 16 32

xy 2log= 0 1 2 3 4 5

Values of x greater than 1

x 21

41

81

161

321

xy 2log= – 1 – 2 – 3 – 4 – 5 Values of x less than 1

The graph of the function xy 2log= is this

Here the y-axis is the vertical asymptote: y goes to negative infinity when x approaches to zero, but never touches the y-axis. 2. Case 0 < a < 1. In this case, like for an exponential function, the function xy alog=

decreases, but now decreases very slowly. Let’s take the example of 21

=a . To draw the

graph of xy21log= without creating a table of values, we can use the result of example

23.3 from the previous session: xx aa

loglog 1 −= . In our particular case,

xx 221 loglog −= .

Let’s consider in general the connection between graphs of the functions )(xfy = and )(xfy −= . Points (x, y) and (x, – y) are symmetrical to each other with respect to the

xy 2log=

193


x-axis. Therefore, the graphs of )(xf and )(xf− are also symmetrical to each other with respect to the x-axis. So, to get the graph of )(xf− , we can just reflect the graph of )(xf with respect to the x-axis. You may recall that we discussed such relationship in session 10 when we compared the graphs of the parabolas 2y x= and 2xy −= .

Applying this reasoning to the function xxy 221 loglog −== , we will get the picture

Here the y-axis remains the vertical asymptote. Relation between Exponential and Logarithmic Functions

If you compare the tables of values of functions xy 2= and xy 2log= , shown above, you may notice that the variables x and y exchange their values. This is not a coincidence. The fact is that the functions xay = and xy alog= are inverse to each other. Let’s consider this concept in general form. Let ( )y f x= be a function. As we already mentioned, we can treat variable x as input that goes into function f, then f operates on x and produces the output y. Schematically, we can represent function f by the diagram A function g is called the inverse to f, if it does the job opposite to f: it passes y back to x. In other words, the input of the inverse function is y, and the output is x. Usually we denote the function inverse to f by 1−f . Schematically, we can represent the inverse function 1−f by the diagram

Note. The notation 1−f for inverse function may create confusion with the notation f1

f x (input) y (output)

f –1 y (input) x (output)

xy21log=

194


for the reciprocal function. Keep in mind that these are completely different functions. To find the function inverse to f, we can solve the equation y = f(x) for x, and then exchange x and y: replace x with y, and y with x.

Example 24.2. Find the inverse function to 2xy = , where 0≥x .

Solution. If we solve the equation 2xy = for non-negative x, we get yx = . Now, just exchange x and y. The inverse function is xy = .

Example 24.3. Find the inverse function to xy alog= .

Solution. As in example 24.2, we solve the equation xy alog= for x. Using the “Circular Rule” described in session 23, we have yax = . Now, exchange x and y, and get the inverse function xay = .

As you see, the logarithmic and exponential functions are inverse to each other. Let’s return to a general case of the function f, and see how the graphs of f and 1−f are related. If (x, f(x)) is a point on the graph of f, then the point ((f(x), x) will be on the graph of 1−f . The points with coordinates (a, b) and (b, a) are symmetrical to each other with respect to the line y = x which is the bisector of the first and third quadrants. (To see that, you may consider some examples, like points (3, 4) and (4, 3), or try to prove this statement in general form). Therefore, the graphs of the function f and its inverse 1−f are symmetrical to each other with respect to the line y = x.

Let’s draw together the graphs of the functions xy 2= and xy 2log= which are inverse to each other:

195


Exercises 24 In exercises 24.1 – 24.8, graphs of the given functions f and g are drawn. One of the graphs is labeled as A and another as B. Match functions f and g with the graphs. Explain your answers. 24.1. ( ) 3 , ( ) 4x xf x g x= = . 24.2. ( ) 4 , ( ) 5x xf x g x= = .

24.3. 1 1( ) , ( )3 4

x x

f x g x = =

. 24.4. 1 1( ) , ( )4 5

x x

f x g x = =

.

24.5. 3 4( ) log ( ), ( ) log ( )f x x g x x= = . 24.6. 4 5( ) log ( ), ( ) log ( )f x x g x x= = .

A

B

B

A

A

A

B

B

A

B

B

A

196


24.7. 1 13 4

( ) log ( ), ( ) log ( )f x x g x x= = . 24.8. 1 14 5

( ) log ( ), ( ) log ( )f x x g x x= = .

In exercises 24.9 and 24.10, find the inverse function to the function f(x).

24.9. a) ( ) 3xf x = 24.10. a) 3( )4

x

f x =

b) 23

( ) log ( )f x x= b) 5( ) log ( )f x x=

In exercises 24.11 and 24.12, find the domains of the given functions. 24.11. a) log (2 3)ay x= + 24.12. a) log (4 5)ay x= + b) log (6 4 )ay x= − b) log (10 8)ay x= −

A A B

B

197

Session 25: Compound Interest and Number e

Session 25

Compound Interest and Number e If you deposit money into a bank, bank pays you interest for usage of your money. If you borrow money from the bank, you pay interest to the bank. The interest may be simple or compound. To explain these, let’s start with some terminology and notations. P – Principal or Initial Value. This is the amount of money you deposit to a bank or borrow from the bank. T – Time. This is the period of time during which the money is used (by you or the bank).

In calculations, it is usually counted in years. If the time period is several months, it is given by fraction or decimal. For example, T = 0.5 means 6 months.

R – Rate. This is the interest rate used to pay for the use of money. This is usually given as percentage per year. In calculations it is used as a decimal. For example, if interest rate is 1.7%, in calculations it is used as 0.017.

I – Interest. This is the amount of money you (or the bank) earn for using money for T years. (Do not confuse interest and rate: interest is the amount in dollar, while rate is the percentage).

A - Amount or Future Value. This is the amount of money that you will have in T years. Obviously, the future value is the sum of two parts: Initial Value and Interest. So, A = P + I.

Simple Interest

This type of interest is usually used when you keep money for a short period of time, for example, several months. This interest is really simple to calculate. If you deposit P dollars for one year at the rate of R (in decimal), then the interest I (this is what you earn), will be equal to I = PR. If you keep money for T years, the total interest earned by you will be I =PRT. This is a formula of simple interest. As you see, it’s really simple. We can also calculate the future value A: A = P + I = P + PRT = P(1 + RT). So, the basic formulas for simple interest are Note. When using the above formulas, keep in mind that the rate R must be taken as a decimal (not as a percent), and time T should be in the same units of time as rate R (usually in years). Example 25.1. Suppose you deposit $800 for 3 months into a bank that pays 5% of simple interest. Calculate the interest that the bank will pay you and the future value (the amount that you withdraw) after 3 months. Solution. We have

P = 800, R = 5% = 0.05, 25.0years123months3 ===T .

I = PTR, A = P(1+RT)

198


Using the above formulas for I and A, we get 1025.005.0800 =××== PRTI (dollars), A = P + I = 800 + 10 = 810 (dollars).

Compound Interest

If you keep money in a bank for a long period of time (for example on CD – Certificate of Deposit, for several years), it is unfair to calculate interest using the above formula for simple interest. Indeed, if the principal is P, and the bank rate is R, then the amount 1A after the first year, according to the formula for future value with T = 1, is )1(1 RPA += . Assume that after the first year you do not withdraw your money. Then for the second year it would be unfair to take as a principal the original value P. Instead, it is reasonable to take the value of 1A (which is, of course, greater than P) as a new principal. In other words, for the second year, the rate R should be applied not only to the initial deposit P, but also to the interest I = PR that you earned for the first year. Notice that according to the formula )1(1 RPA += , the amount at the end of a year is equal to the amount at the beginning of the year times )1( R+ . Therefore, at the end of the second year the amount, denoted as 2A , should be

212 )1()1)(1()1( RPRRPRAA +=++=+= .

If we continue the same reasoning, then in T years you will accumulate the amount of T

T RPA )1( += dollars. Formula

TRPA )1( +=

allows us to calculate the future value after T years, if the initial principal is P, and the bank interest is R.

Although the above formula for future value is fairer than the corresponding formula for simple interest, it is still not fair enough. Here is the reasoning. After a certain (even short) period of time, say, in half a year your principal will be increased by earned interest for this period. However, according to the above formula, the original principal remains unchanged throughout the year, and only at the beginning of the next year the bank recalculates, and replaces the original principal with a new value. It would be better (for customers), if such recalculates were done more often. Many banks do that. They introduce a parameter called the compounding period. This is the period of time after which the bank recalculates the principal: the bank takes the principal, adds the earned interest and uses this sum as the new principal. Usually, bank compounds (recalculates) semiannually (every half of a year), quarterly (every three months), monthly, and even daily. Therefore, the above formula for the future value should be modified by including a new parameter N – the number of compounding periods per year. If interest is compounded yearly, then N = 1; if semi-annually, then N = 2; quarterly, then N = 4; monthly, then N = 12; weekly, then N = 52; daily, then N = 365. Let’s modify the above formula for future value if investment compounded monthly, i.e. N = 12. Since the rate R is constant throughout the year, interest for one month will be

199


12PRI = . After the 1st month, future value is 1

12 12PR RA P I P P = + = + = +

. So, in

order to get the future value for any month, we should take the future value for the

previous month and multiply it by the expression 112R

+ . Therefore, after the 2nd month,

the future value is

2

1 1 112 12 12R R RA P P = + + = +

.

And, at the end of the year T, the future value becoms12

112

TRA P = +

. In similar way,

for any compound period N, we can get the general compound interest formula:

.

Here RN

is the rate for one compounding period, and TN is the total number of

compounding periods for T years. For example, if rate R = 2.4%, deposit is compounded

quarterly (N = 4), and number of years T = 5, then 0.024 0.0064

RN= = and

5 4 20TN = ⋅ = . Interest I on this deposit is the difference between future value A and the original principal P:

−

+=−

+=−= 111

TNTN

NRPP

NRPPAI .

Example 25.2. Suppose you deposit $300 for 8 years at 3% compounded quarterly. Find the future value and earned interest.

Solution. We have: P = 300, R = 3% = 0.03, N = 4, T = 8. Substitute these values into the compound interest formula and calculate future value A:

8 4

320.03300 1 300 (1.0075) 300 1.2701 381.034

A⋅

= ⋅ + = ⋅ ≈ ⋅ =

.

So, the future value is $381.03. Interest I is the difference:

I = A – P = 381.03 – 300 = 81.03.

Therefore, you will earn $81.03 for 8 years.

TN

NRPA

+= 1

200


The compound interest formula can be used to find the rate R, or time T needed to accumulate the desired amount in the future. Let’s solve the problem to find time T in general form. Dividing both parts of the compound interest formula by P, we have

TN

NR

PA

+= 1 .

Now, take log from both sides:

+⋅=

+=

NRTN

NR

PA TN

1log1loglog .

From here,

+⋅

=

NRN

PA

T1log

log.

Note. You do not need to memorize this formula. To find time in particular problems, plug in given data into the compound interest formula, and apply log to both sides.

Example 25.3. Suppose you deposit some amount of money at 6% compounded monthly. In how many years your deposit will be doubled?

Solution. According to the problem, the future value A is twice as the principal P: A = 2P. Also, R = 6% = 0.06, N = 12. Substitute these data into the compound interest formula:

120.061 2 112

TN TRA P P PN

= + ⇒ = +

.

Reduce (divide) both side by P and calculate expression inside parentheses: ( )122 1.005 T= . Now apply log to both sides

( ) ( )12log 2 log 1.005 log 2 12 log 1.005T T= ⇒ = ⋅ ⋅ .

From here log 2 11.612 log(1.005)

T = =⋅

. . So, your deposit will be doubled in about 11.6 years. When you decide in which bank to deposit your money or which credit card to use to make only minimum payments, you need to consider not only the rate, but also the compounding period. To make a true comparison of different rates, we can compare the interest that accrues on one dollar for one year. This value is called the effective rate or APY (Annual Percentage Yield). To get a formula for APY, we substitute the values P = 1 and T = 1 into the formula for interest I (formula is above example 22.2). We will have

201


11 −

+=

N

NRAPY .

Usually, APY is presented as percentage.

Example 25.4. Suppose you have the choice of using two credit cards for which you want to make only minimum payments. On the 1st card, you will pay 18% interest compounded monthly, and on the 2nd card – 17. 9% compounded daily. Which deal is best for you? Solution. On the first glance, it looks like the 2nd card is better (you pay smaller rate). However, let’s compare APYs for these two cards.

1st card: 1956.011218.01

12

≈−

+=APY , or APY = 19.56%.

2nd card: 1960.01360179.01

360

≈−

+=APY , or APY = 19.60%.

As you can see, although the rate on the 1st credit card is higher, you would prefer this card because its APY is lower and in the long run you will pay less interest.

Continuous compound interest. Number e

We saw that when you invest in a bank, it’s more profitable for you, if bank uses the compound interest formula instead of the simple one. Also, the shorter the compounding period, the greater your profit. We mentioned the cases of compounding semiannually, quarterly, monthly, and daily. But why should we be limited only with to these periods? Can the compounding period to be one hour, one minute, or even one second? The answer is yes. In this way we come up to a formula called continuous compound interest. To get this formula, let’s modify the compound interest formula:

TR

RNTN

NRP

NRPA

+=

+= 11 .

If we denote RN

N NRe

+= 1 , then ( )TR

NePA = .

Let’s see what happens to your income when the compounding period becomes shorter and shorter. In this case, interest is recalculated more often, and, as a result, the future value (i.e. your money) becomes larger. You might think that the future value will grow endlessly and, eventually, can become huge. However (unfortunately for you) this is not so. Your income has a limit. If N increases to infinity (becomes bigger and bigger), then

202


TRPeA =

the value RN also increases to infinity. However, this is not the case for Ne (and also for

future value). This value also increases but not to infinity. Let’s calculate Ne for some

values of RN .

RN 1 2 3 4 5 10 100 1000

Ne 2 2.25 2.37 2.44 2.49 2.59 2.70 2.717 It can be shown (it is done in Calculus courses) that if N continues to increase, the value of Ne cannot be greater than 3. In fact, Ne is getting closer and closer to a certain constant number. This number is denoted by the letter e and is approximately equal to 2.718. This number is called the Euler's number or the base of the natural logarithms. Most scientific calculators have a button to calculate the number e with even greater accuracy.

Let’s return to the formula RN

N NRe

+= 1 and denote power N n

R= . Then, by taking the

reciprocal, 1RN n= and 11

n

Nen

= +

. If N is big, number n is also big and Ne e≈ . We

may say that

11 2.718n

en

≈ + ≈

, if number n is big.

If we replace Ne in the formula ( )TRNePA = with the number e, we will get the

continuous compound interest formula This formula gives the maximum possible future value compared to the compound interest formula with any finite number N of compounding periods per year. Example 25.5. Suppose you invested $300 for 8 years at 3% compounded continuously. Find the future value and earned interest.

Solution. We have: P = 300, R = 3% = 0.03, T = 8. Substitute these values into the above formula and calculate future value A:

8 0.03 0.24300 300 300 1.27125 381.37TRA Pe e e⋅= = ⋅ = ⋅ = ⋅ =

So, the future value is $381.37. Interest I is the difference:

I = A – P = 381.37 – 300 = 81.37. Compare this result with the result of example 25.2. The difference in interest is small (34 cents), but for large investments the difference becomes significant.

203


Exercises 25

Round all answers to the nearest cent. 25.1. Suppose you deposit $700 for 5 months into a bank that pays 2.5% simple interest.

Calculate interest that bank will pay you and future value (amount that you withdraw) after 5 months.

25.2. Suppose you deposit $600 for 7 months into a bank that pays 4% simple interest. Calculate interest that bank will pay you and future value (amount that you withdraw) after 7 months.

25.3. Suppose you deposit $1,200 for 5 years at 5% compounded monthly. Find the future value and earned interest.

25.4. Suppose you deposit $2,400 for 6 years at 2% compounded semiannually. Find the future value and earned interest.

25.5. Suppose you made an investment of $1,200 for 5 years at 5% compounded continuously. Find the future value and earned interest. Compare with the results of example 25.3.

25.6. Suppose you made an investment of $2,400 for 6 years at 2% compounded continuously. Find the future value and earned interest. Compare with the results of example 25.4.

25.7. Suppose you deposit a certain amount of money at 4% compounded quarterly. In how many years will your deposit be doubled?

25.8. Suppose you deposit a certain amount of money at 7% compounded daily. In how many years will your deposit be doubled?

Challenge Problem

25.9. Suppose you want to invest $5,000 for the next 10 years. You are considering two banks for your investment. Bank A offers a rate of 7 % compounded semiannually. Bank B offers a rate of 6.9 % compounded daily. Which bank would you choose? Is it possible to solve this problem without information on the investment amount and time?

204

Answers to Exercises


Session 1 1.1. a) System has unique solution ( )5, 2, 1− −

b) System has unique solution ( )2, 1, 3− − .

c) System is dependent. Possible parametric form: 1, 2 ,2

x t y t z t= = − = .

Particular solutions for 0t = : 10, , 02

⎛ ⎞⎜ ⎟⎝ ⎠

.

d) System has unique solution ( )0, 2, 1−

e) System is inconsistent: no solutions.

1.2. a) System has unique solution ( )4, 2, 3−

b) System has unique solution ( )3, 4, 2− − .

c) System is inconsistent: no solutions.

d) System has unique solution ( )1, 0, 2

e) System is dependent. Possible parametric form: 1 , , 1x t y t z t= − = = + . Particular solutions for 0t = : ( )1, 0, 1 .

Session 1A

1A.1. 64 52,17 17⎛ ⎞⎜ ⎟⎝ ⎠

1A.2. 55 17,23 23

⎛ ⎞−⎜ ⎟⎝ ⎠

1A.3. 22D = 1A.4. 6D = −

1A.5. 35 37 9, ,22 22 22

⎛ ⎞−⎜ ⎟⎝ ⎠

1A.6. 11 17 29, ,6 3 6

⎛ ⎞⎜ ⎟⎝ ⎠

Session 2 2.1. a) 23 14 0; 3, 1, 14x x a b c+ − = = = = −

b) 216 24 3 0; 16, 24, 3x x a b c− + = = = − =

2.2. a) 28 2 3 0; 8, 2, 3x x a b c− − = = = − = −

b) 225 20 1 0; 25, 20, 1x x a b c+ + = = = =

2.3. a) 7 , 23

⎧ ⎫−⎨ ⎬⎩ ⎭

b) 56⎧ ⎫⎨ ⎬⎩ ⎭

2.4. a) 3 1,4 2

⎧ ⎫−⎨ ⎬⎩ ⎭

b) 47

⎧ ⎫−⎨ ⎬⎩ ⎭

2.5. a) 50,3

⎧ ⎫−⎨ ⎬⎩ ⎭

b) { 4}± c) { 4, 3}− d) { 5}− e) {4, 5}−

205


Session 2 (continued)

2.6. a) 70,6

⎧ ⎫⎨ ⎬⎩ ⎭

b) { 3}± c) { 3, 5}− d) { 6}− e) {6, 4}−

2.7. a) 3 1,4 2

⎧ ⎫−⎨ ⎬⎩ ⎭

b) 3 , 35⎧ ⎫−⎨ ⎬⎩ ⎭

c) 1 2,3 3⎧ ⎫⎨ ⎬⎩ ⎭

2.8. a) 1 5,3 2⎧ ⎫−⎨ ⎬⎩ ⎭

b) 4 , 17⎧ ⎫−⎨ ⎬⎩ ⎭

c) 1 2,4 3

⎧ ⎫−⎨ ⎬⎩ ⎭

Session 3

3.1. 42.5 10× 3.2. 42.9 10× 3.3. 42.56 10−× 3.4. 51.4 10−× 3.5. 0.00004 3.6. 0.0875 3.7. 3,475,000 3.8. 1,236 3.9. a) 33.47 10−× , b) 22.5 10× 3.10. a) 54.38 10−× , b) 43.6 10× 3.11. a) 2, b) 1, c) 1/16, d) – 1/16, e) 1/16 3.12. a) 3, b) 1, c) 1/25, d) – 1/25, e) 1/25 3.13. a) 21/ a , b) 1/ c6 , c) 1/ a6 , d) n2 /m2 3.14. a) 31/ c , b) 61/ n , c) 121/ d , d) 3 3/b a− 3.15. a) 4p , b) 121/ p , c) 12p , d) 41/ p , e) 41/ p , f) 12p , g) 121/ p , h) 4p 3.16. a) 6r , b) 12r , c) 121/ r , d) 61/ r , e) 61/ r , f) 121/ r , g) 12r , h) 6r 3.17. /b d a cy x+ + 3.18. /w y x zp q+ + 3.19. /mnb a 3.20. 2 3/xyu v 3.21. ( ) ( )30 1636 / 25s r 3.22. ( ) ( )60 36125 / 8x y− 3.23. ( ) ( )18 218 / 27p q− 3.24. ( ) ( )20 89 / 25m n

Session 4

4.1. 6 45y −

4.2. 12 87z −

4.3. ( )( )3 52 3z zz+

− 4.4.

( )2 2 33 5y yy

+

−

4.5. ( )( )2 53 4xx+

+ 4.6.

( )( )3 24 1xx−

− 4.7.

( )2 47xx+

4.8. ( )3 34xx+

206



4.9. 2 4.10. 2 4.11. 10 1536m+

4.12. 7 5060n +

4.13.

14 1524x yxy+

4.14. 9 2824x yxy

+ 4.15.

2

2

35 50 830x x

x− +

4.16. 2

2

9 28 1024

x xx

− +

4.17.

( )( )23

4 7 7 4x

x x+

− − 4.18.

( )( )3 10

3 4 4 3x

x x+

− −

4.19.

24a

a −

4.20.

25b

b −

4.21.

32x

x − 4.22.

24x

x − 4.23.

( )3

5 5xx+

− 4.24.

( )5

4 4xx+

−

4.25.

( )( )23 5x x+ +

4.26. ( )( )

14 5x x+ +

4.27. ( )( )

134 2b b

−+ −

4.28. ( )( )

253 5c c

−− +

4.29. 2

49dc

4.30. 3

15nm

4.31. 3 24 5xx

+

− 4.32.

2 35 4xx

−

+

4.33. 143x

4.34. 154x

4.35. 724

− 4.36. 412

4.37. 2

4 36 2x xyy x+

− 4.38. 2

6 54 3xy yy x−

+

4.39. – 3

4.40. – 4

4.41.

4 196 13kk−

− 4.42.

5 127 34mm−

− 4.43.

14

xx−

+ 4.44.

12

xx−

+

Session 5

5.1. 9 / 7x = 5.2. 5x = 5.3. 4 / 9x = 5.4. 6 / 5x = − 5.5. 17 / 3x = 5.6. 7x = 5.7. 23x = 5.8. 17x = 5.9. 6 / 5x = 5.10. 25 / 2x = 5.11. 2 / 9x = 5.12. 1/ 6x = 5.13. 1/12x = − 5.14 11/ 5x = − 5.15. 11x = − 5.16. 6x = − 5.17. 113/11m = − 5.18. 15m = − 5.19. 3x = 5.20. 11x = −

5.25. 2 gallons 5.26. 4 gallons 5.27. 36% 5.28. 14%

Session 6

6.1. a) 4 b) 3 c) 25 d) 4 e) 1/2 f) 125 g) 1/9 h) 2 6.2. a) 9 b) 4 c) 36 d) 3 e) 1/6 f) 4 g) 1/8 h) 3

207



6.3. 1a

6.4. b 6.5. 5 2 6.6. 3 6

6.7. 4 2 6.8. 6 2 6.9. a) 4x

b) 3x x 6.10. a) 3x b) 2x x

6.11. a) 186y

b) 43y y 6.12. a) 247z z b) 328z

6.13. 5 2 815 3x y z y 6.14. 7 4 625 3x y z yz

6.15. ( ) ( )16 9/b a 6.16. ( ) ( )9 25/a b

Session 7

7.1. a) 7 b) 2018 c) 47 2 7.2. a) 5 b) 2019 c) 26 3 7.3. 7 6 7.4. 6 15

7.5. a) 4 43 5p q q b) 5576 3a b b 7.6. a) 5 45 3m n m b) 6 3294 2u v v

7.7. a) 11 6 b) 2 7− 7.8. a) 9 5 b) 3 3− 7.9. 8 6 2 7m n k+

7.10. 12 3 2 6p q r+ 7.11. a) 7 2 b) 7 3 c) 10 6 10 5−

7.12. a) 5 3 b) 2 5 c) 28 7 40 3− 7.13. 2 5 5 2+ 7.14. 2 3 3 2+

7.15. 315 7.16. 24 7.17. 4 8 30+ 7.18. 26 2 14− − 7.19. 42 7.20. 67

7.21. 117 7.22. 189 7.23. 83 12 35+ 7.24. 72 48 2+ 7.25. 83 12 35−

7.26. 72 48 2− Session 8

8.1.

5 66

8.2. 4 33

8.3. 155

8.4. 427

8.5. 515

8.6. 315

8.7. 4 315

8.8. 2 515

8.9. 2 5− 3( )11

8.10. 3 52−

8.11. 2 3+ 8.12.

5 619−

8.13.

6 33−

8.14. 6 5+ 8.15. 4 3 8 6

16 9u uv u v

u v+ + +

−

208



8.16. 2 3 54 25a ab ba b− −

−

8.17. 7x =

8.18. 13x =

8.19. 8x =

8.20. 5x = 8.21. No solutions 8.22. No solutions 8.23. 4x = 8.24. 3x = 8.25. No solutions 8.26. No solutions

8.27. a) x = 2 b) 2x = − and 1x = − 8.28. a) 3x = − and 1x = − b) 1x =

Session 9 9.1. a) 5i

b) 4 2i

9.2. a) 4i

b) 3 3i

9.3. a) 9 2i−

b) 4 8i− −

9.4. a) 12 5i−

b) 4 5i+

9.5. a) 6 12i− b) 8 20i− + c) 7 22i− + d) 12 54i− − e) 45 7i− f) 13

9.6. a) 12 20i+ b) 21 42i+ c) 4 38i+ d) 17 57i− e) 34 38i− f) 20

9.7. a) 8 25 5

i+ b) 32i− c)

3 92 4

i+ d) 38 961 61

i− − e) 2 95 5

i− f) 5 1213 13

i−

9.8. a) 732i− + b)

23i c)

1 22 3

i− d) 38 925 25

i− e) 2 2313 13

i− − f) 45 2853 53

i− −

9.9. a) 1− b) i− c) 1 d) i

9.10. a) i− b) 1 c) i d) 1−

9.11. a) 3 64

⎧ ⎫±⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

b) { }7i± c) 2 33 6

i⎧ ⎫⎪ ⎪− ±⎨ ⎬⎪ ⎪⎩ ⎭

9.12. a) 2 35

⎧ ⎫− ±⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

b) { }3i± c) 5 67 7

i⎧ ⎫⎪ ⎪

±⎨ ⎬⎪ ⎪⎩ ⎭

Session 10

10.1. a) ( )22 8 16 4x x x+ + = + b) 2

2 9 334 2

x x x⎛ ⎞− + = −⎜ ⎟⎝ ⎠

c) 2

2 1 14 2

x x x⎛ ⎞+ + = +⎜ ⎟⎝ ⎠

d) 2

2 5 25 54 64 8

x x x⎛ ⎞− + = −⎜ ⎟⎝ ⎠

209



10.2. a) ( )22 6 9 3x x x+ + = + b) 2

2 81 994 2

x x x⎛ ⎞− + = −⎜ ⎟⎝ ⎠

c) 2

2 1 14 2

x x x⎛ ⎞− + = −⎜ ⎟⎝ ⎠

d) 2

2 7 49 73 36 6

x x x⎛ ⎞+ + = +⎜ ⎟⎝ ⎠

10.3. a) { }1, 5− b) { }3, 2− − c) 7 172

⎧ ⎫±⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

d) 33,2

⎧ ⎫−⎨ ⎬⎩ ⎭

10.4. a) { }4, 2− b) { }2, 7 c) 3 292

⎧ ⎫− ±⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

d) 12,3

⎧ ⎫− −⎨ ⎬⎩ ⎭

10.5. a) No real roots b) Two roots c) One root 10.6. a) One root b) No real roots c) Two roots

10.7. a) 4 215

⎧ ⎫±⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

b) 34

⎧ ⎫−⎨ ⎬⎩ ⎭

c) 7 318 8

i⎧ ⎫⎪ ⎪

±⎨ ⎬⎪ ⎪⎩ ⎭

d) 4 3,3 2

⎧ ⎫⎨ ⎬⎩ ⎭

10.8. a) 3 307

⎧ ⎫±⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

b) 73

⎧ ⎫⎨ ⎬⎩ ⎭

c) 5 4712 12

i⎧ ⎫⎪ ⎪− ±⎨ ⎬⎪ ⎪⎩ ⎭

d) 5 3,6 2⎧ ⎫⎨ ⎬⎩ ⎭

Session 11

11.1. a) ( )4, 5− − , Up b) ( )2, 6 , Down c) ( )5, 4− , Up

11.2. a) ( )3, 7 , Down b) ( )7, 3− − , Up c) ( )1, 8− , Down

11.3. a) squared form: ( )23 2 3y x= + − , standard form: 23 12 9y x x= + +

b) squared form: ( )23 2 1y x= + + , standard form: 23 12 13y x x= + +

11.4. a) squared form: ( )23 2 1y x= − + , standard form: 23 12 13y x x= − +

b) squared form: ( )23 2 3y x= − − , standard form: 23 12 9y x x= − +

11.5. a) squared form: ( )22 3 2y x= − − + , standard form: 22 12 16y x x= − + −

b) squared form: ( )22 3 1y x= − − − , standard form: 22 12 19y x x= − + −

11.6. a) squared form: ( )22 3 1y x= − + − , standard form: 22 12 19y x x= − − −

b) squared form: ( )22 3 2y x= − + + , standard form: 22 12 16y x x= − − −

210



11.7. a) vertex: ( )1, 4− , y- intercepts: ( )0, 3− , x-intercepts: ( )1, 0− and ( )3, 0

b) vertex: ( )2, 9− , y- intercepts: ( )0, 5 , x-intercepts: ( )5, 0− and ( )1, 0

11.8. a) vertex: ( )1, 9− , y- intercepts: ( )0, 8 , x-intercepts: ( )4, 0− and ( )2, 0

b) vertex: ( )3, 4− , y- intercepts: ( )0, 5 , x-intercepts: ( )1, 0 and ( )5, 0

Session 12 12.1. 2 10 12.2. 5 2

12.3. Acute triangle. Biggest angle is C, smallest angle is A 12.4. Right triangle. Biggest angle is A, smallest angle is C

12.5. ( )1, 5− − 12.6. ( )3, 1 12.7. ( )1, 3 12.8. ( )7, 2− −

12.9. a) Center ( )2, 4− − , radius = 6 b) Center ( )5, 2− , radius = 2 5

12.10. a) Center ( )6, 3 , radius = 7 b) Center ( )7, 8− , radius = 5 2

12.11. ( ) ( )2 2 23 4 13x y+ + + = 12.12. ( ) ( )2 2 212 10 17x y− + − =

12.13. a) Center ( )2, 1− , radius = 4, points: ( ) ( ) ( ) ( )2, 5 , 2, 3 , 6, 1 , 2, 1− − − −

b) Center ( )5, 4 , radius = 3, points: ( ) ( ) ( ) ( )5, 1 , 5, 7 , 2, 4 , 8, 4

12.14. a) Center ( )4, 1− , radius = 3, points: ( ) ( ) ( ) ( )4, 2 , 4, 4 , 1, 1 , 7, 1− − −

b) Center ( )3, 2− − , radius = 5, points: ( ) ( ) ( ) ( )3, 7 , 3, 3 , 8, 2 , 2, 2− − − − − −

Session 13

13.1. a) ( ) ( ){ }7, 22 , 1, 2− − 13.2. a) ( ) ( ){ }1, 6 , 13, 50− −

b) ( ){ }8, 3− − b) ( ){ }70, 6

c) ( ){ }9, 3− c) ( ){ }4, 2

d) ( ) ( ){ }19, 11 , 1, 1− − − d) ( ) ( ){ }34, 12 , 2, 0−

e) ( ) ( ) ( ) ( ){ }3, 4 , 3, 4 , 3, 4 , 3, 4− − − − e) ( ) ( ) ( ) ( ){ }2, 4 , 2, 4 , 2, 4 , 2, 4− − − −

f) ( ) ( ) ( ) ( ){ }1, 3 , 1, 3 , 3, 1 , 3, 1− − − − f) ( ) ( ) ( ) ( ){ }2, 5 , 2, 5 , 2, 5 , 2, 5− − − −

13.3. Length = 4 m, width = 3 m 13.4. Length = 6 yd, width = 5 yd

211


Session 14

14.1. a) 50 360 , 0, 1, 2,...n n+ = ± ±o o b) 70 360 , 0, 1, 2,...n n− + = ± ±o o 410o, 770o, 310− o, 670− o 290o, 650o, 430− o, 790− o

14.2. a) 27 360 , 0, 1, 2,...n n+ = ± ±o o b) 35 360 , 0, 1, 2,...n n− + = ± ±o o

387o , 747o, 333− o, 693− o 325o, 685o, 395− o, 755− o

14.3. a) 6 3, 12b c= = b) 3, 2 3a c= = c) 4, 4 3a b= =

14.4. a) 8 3, 16b c= = b) 3 3, 6 3a c= = c) 2, 2 3a b= =

14.5. a) 4, 4 2b c= = b) 8, 8 2a c= = c) 9 2 / 2a b= =

14.6. a) 3, 3 2b c= = b) 6, 6 2a c= = c) 7 2 / 2a b= =

14.7. Side opposite to 30o is / 2c , side opposite to 60o is 3 / 2c

14.8. Side opposite to 30o is 3 / 3b , hypotenuse is 2 3 / 3b

14.9. Both sides are 2 / 2c Session 15

15.1. 3 4 3 4 3 4sin , cos , tan , sin , cos , tan5 5 4 5 5 3

A A A B B B= = = = = = .

15.2. 12 5 12 5 12 5sin , cos , tan , sin , cos , tan13 13 5 13 13 12

A A A B B B= = = = = = .

15.3. 2 3 2 3sec30 , csc30 2, sec45 csc45 2, sec60 2, csc603 3

= = = = = =o o o o o o .

15.4. 3cot 30 3, cot 45 1, cot 603

= = =o o o . 15.5. B 15.6. A

15.7. A 15.8. B 15.9. 29.1 ft 15.10. 61 ft 15.11. 11.5 ft

15.12. 2.5 m 15.13. 167.1 ft 15.14. 83.1 ft 15.15. 3.1o 15.16. 29.1o

15.17. 46.1o 15.18. 21.1o 15.19. 63.9o 15.20. 53.1o 15.21. 20.3 m

15.22. 60o Session 16

16.1. a) sinθ− b) sinθ 16.2. a) cosθ− b) cosθ−

16.3. a) 50o b) 40o c) 70o d) 85o e) 50o

16.4. a) 20o b) 10o c) 50o d) 80o e) 30o

212


Session 16 (continued) 16.5. a) 220o b) 110o c) 310o d) 20o 16.6. a) 320o b) 70o c) 130o d) 200o

16.7. a) III, 130 ,2

−o b) IV, 160 ,2

o c) II, 45 , 1−o

16.8. a) IV, 245 ,2

−o b) II, 330 ,2

−o c) III, 60 , 3o

16.9.

a) IV, 5 2 5 3 3 5 5cos , tan , csc , sec , cot3 5 2 5 2

θ θ θ θ θ= = − = − = = −

b) II, 21 21 5 21 5 2 21sin , tan , csc , sec , cot5 2 21 2 21

θ θ θ θ θ= = − = = − = −

c) III, 3 34 5 34 34 34 5sin , cos , csc , sec , cot34 34 3 5 3

θ θ θ θ θ= − = − = − = − =

16.10.

a) II, 33 4 33 7 7 33 33cos , tan , csc , sec , cot7 33 4 33 4

θ θ θ θ θ= − = − = = − = −

b) III, 39 39 8 39 8 5 39sin , tan , csc , sec , cot8 5 39 5 39

θ θ θ θ θ= − = = − = − =

c) IV, 7 65 4 65 65 65 4sin , cos , csc , sec , cot65 65 7 4 7

θ θ θ θ θ= − = = − = = −

16.11.

a) 2 5 5 5 1sin , cos , tan 2, csc , sec 5, cot5 5 2 2

θ θ θ θ θ θ= − = − = = − = − =

b) 5 41 4 41 5 41 41 4sin , cos , tan , csc , sec , cot41 41 4 5 4 5

θ θ θ θ θ θ= − = = − = − = = −

c) 7 58 3 58 7 58 58 3sin , cos , tan , csc , sec , cot58 58 3 7 3 7

θ θ θ θ θ θ= = − = − = = − = −

16.12.

a) 5 34 3 34 5 34 34 3sin , cos , tan , csc , sec , cot34 34 3 5 3 5

θ θ θ θ θ θ= = − = − = = − = −

b) 6 61 5 61 6 61 61 5sin , cos , tan , csc , sec , cot61 61 5 6 5 6

θ θ θ θ θ θ= = − = − = = − = −

c) 7 65 4 65 7 65 65 4sin , cos , tan , csc , sec , cot65 65 4 7 4 7

θ θ θ θ θ θ= − = − = = − = − =

213


Session 17

17.1. 105.3, 129.1, 20p q R= = = o 17.2. 10.8, 21.3, 85q r P= = = o

17.3. 21.2, 25.4 , 114.6l K L= = =o o 17.4. 36.4, 77.5 , 32.5k K M= = =o o

17.5. 9.1 m 17.6. 21.9 m 17.7. AB = 33 m, BC = 26.9 m

17.8. 11.1 ft and 12.0 ft. 17.9. 39.9o 17.10. 37.1!

17.11. 98.6o 17.12. 112.3o 17.13. 11.2 m

17.14. 32.5 m 17.23. 3.8 mi 17.24. ( )

sin sinsind A B

A B+

17.25. 6.3 mi 17.26. sin sinsind A B

A B− 17.27. 1.3 m and 2.2 m

Session 18

18.1. 28.2, 40.1 , 74.9p Q R= = =o o 18.2. 96.3, 16.2 , 33.8q P R= = =o o

18.3. 28.9 , 46.6 , 104.5K L M= = =o o o 18.4. 133.4 , 29.0 , 17.6K L M= = =o o o

18.5. Yes 18.6. 301.4 ft 18.7. 41.8o 18.8. 29.3o

18.9. 14o 18.10. 57o 18.11. 11.5o

Session 19

19.1. a) 74.5o, b) 34.4− o 19.2. a) 137.5o, b) 45.8− o

19.3. a) 40o b) 54− o 19.4. a) 48o , b) 216− o

19.5. a) 2.44, b) – 1.48 19.6. a) 1.36, b) – 4.14

19.7. a) 23π

, b) 56π

− 19.8. a) 116π

, b) 54π

−

19.9. a) 4 3 4 1 4sin , cos , tan 33 2 3 2 3π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

b) 2 2sin , cos , tan 1

4 2 4 2 4π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

c) 5 1 5 3 5 3sin , cos , tan6 2 6 2 6 3π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

19.10. a) 1 3 3sin , cos , tan

6 2 6 2 6 3π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

214



b) 2 3 2 1 2sin , cos , tan 33 2 3 2 3π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

c) 3 2 3 2 3sin , cos , tan 14 2 4 2 4π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − = − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

19.11. a) 9.7 ft b) 1.8 cm 19.12. a) 0.5 m b) 5.0 in 19.13. 5, ,

4 3 12π π π

19.14. 16 cabs, arc ≈ 3.14 m 19.15. 180 m/min 19.16. 3 sec

Session 20

20.1. a) 3,

4 4π π

, b) 4 5,3 3π π

20.2. a) 2,

3 3π π

, b) 7 11,6 6π π

20.3. No solutions 20.4. No solutions 20.5. a) 0.85, 2.29 b) 3.48, 5.94 20.6. a) 0.64, 2.50 b) 3.75, 5.67

Session 21

21.1. a) 7,

4 4π π

, b) 5 7,6 6π π

21.2. a) 5,

3 3π π

, b) 3 5,4 4π π

21.3. a) 7,

6 6π π

, b) 2 5,3 3π π

21.4. a) 5,

4 4π π

, b) 5 11,6 6π π

21.5. No solutions 21.6. No solutions 21.7. a) 0.93, 5.35 b) 2.18, 4.10 21.8. a) 0.72, 5.56 b) 1.91, 4.37

21.9. a) 1.11, 4.25 b) 2.68, 5.82 21.10. a) 1.33, 4.47 b) 2.55, 5.70

Session 22

22.13. 30, ,

2 2π π

22.14. 0, ,2π

π 22.15. 30, ,2π

π

22.16. 3, ,

2 2π π

π 22.17. 3 5 7, , ,

4 4 4 4π π π π

22.18. 3 5 7, , ,

4 4 4 4π π π π

22.19. 7 11,6 6π π

22.20. 2 4,3 3π π

22.21. , 0.841, 5.442π

22.22. , 4.069, 5.3562π

22.23. 70, , ,

6 6π π

π 22.24. 4 3, , ,

3 2 3 2π π π π

22.25. ,2π

π 22.26. 30,2π

215


Session 23

23.1. a) 42 16= b) 4 1216

− = 23.2. a) 34 64= b) 3 1464

− =

23.3. a) 2log 8 3= b) 21log 38= − 23.4. a) 3log 81 4= b) 3

1log 481

= −

23.5. a) 4 b) 3 c) – 2 d) – 6 e) – 2 f) 15

g) 94

23.6. a) 2 b) 4 c) – 3 d) – 3 e) – 2 f) 14

g) 115

23.7. a) 3 b) 23

− c) – 3 23.8. a) 4 b) 12

c) – 2

23.9. a) 7 b) 4 23.10. a) 4 b) 3 23.11. 1 23.12. 3 23.13. 3u + 2v

23.14. 4u + 5v 23.15. 7u – 6v 23.16. 3v – 2u 23.17. 1.069 23.18. 0.936

23.19. a) 1.226 b) – 0.429 c) – 0.348 23.20. a) 1.292 b) 0.881 c) 1.931

Session 24

24.1. ,f B g A↔ ↔ 24.2. ,f A g B↔ ↔ 24.3. ,f A g B↔ ↔

24.4. ,f B g A↔ ↔ 24.5. ,f B g A↔ ↔ 24.6. ,f A g B↔ ↔

24.7. ,f A g B↔ ↔ 24.8. ,f B g A↔ ↔ 24.9. a) 3log ( )x b) 23

x⎛ ⎞⎜ ⎟⎝ ⎠

24.10. a) 34

log ( )x b) 5x 24.11. a) ( )3/ 2,− ∞ b) ( ), 3/ 2−∞

24.12. a) ( )5/ 4,− ∞ b) ( ), 5 / 4−∞

Session 25

25.1. I = $7.29, A = $ 707.29 25.2. I = $14.00, A = $ 614.00

25.3. A = $1,540.03, I = $ 340.03 25.4. A = $2,704.38, I = $ 304.38

25.5. A = $1,540.83, I = $ 340.83 25.6. A = $2,705.99, I = $ 305.99

25.7. 17.4 years 25.8. 9.9 years

25.9. Bank B. Yes, information about investment amount and time is not needed.

216

college algebra and trigonometry college algebra and trigonometry by alexander rozenblyum is...

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